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Lecture Comments (2)

1 answer

Last reply by: Professor Eric Smith
Mon May 18, 2015 12:56 PM

Post by Lee Ross on May 18, 2015

Hi Eric,

Example 2. Just so I'm following... the second step you have written out is 2Xsqr +4X = 2. You don't combine the 2xsqr + 4x because they are not like terms? What if it was 2Xsqr + 4Xsqr = 2?

Solving Quadratic Equations

  • The principle of square roots says if we have x2 = k, then x = √k or x = -√k. Note this splits it into two numbers. This method works well if the only variable is being squared.
  • To complete the square
    • Isolate the variables on one side of the equal sign
    • Divide all terms by the coefficient of the squared term
    • Find the number to add to both sides of the equation. This is done by dividing the coefficient on the middle term by 2, and then squaring it.
    • Factor the polynomial
    • Use the principle of square roots when taking the square root of both sides
    • Solve each problem for the variable separately
  • The quadratic formula can be built from completing the square. Because of this it can be used to solve all quadratic equations. Make sure to property identify the coefficients used for a, b, and c.
  • The discriminant can be used to determine the types of solution the quadratic equation has. Remember the discriminant is the part of the quadratic formula underneath the square root.
  • Factoring is often the quickest method, but the quadratic formula will also work. Be familiar with the many ways of solving a quadratic so that you can always choose the best method for a particular quadratic equation.

Solving Quadratic Equations

x2 + 4x + 18 = 2
  • x2 + 4x = − 16
  • x2 + 4x + ( [b/2] )2 = − 16 + ( [b/2] )2
  • x2 + 4x + [16/4] = − 16 + [16/4]
  • x2 + 4x + 4 = − 16 + 4
  • (x + 2)2 = − 12
x + 2 = ±√{12}
No Solution
y2 + 10y + 14 = 70
  • y2 + 10y = 56
  • y2 + 10y + ( [b/2] )2 = 56
  • y2 + 10y + [100/4] = 56 + 25
  • ( y + 5 )2 = 81
  • y + 5 = √{81}
  • y + 5 = ±9
  • y + 5 = 9y = 4
  • y + 5 = − 9y = − 13
y = { − 13,4}
k2 − 14k − 27 = 68
  • k2 − 14k = 95
  • k2 − 14k + ( [b/2] )2 = 95 + ( [b/2] )2
  • k2 − 14k + [196/4] = 95 + [196/4]
  • k2 − 14k + 49 = 95 + 49
  • ( k − 7 )2 = 144
  • k − 7 = ±12
  • k − 7 = 12k = 19
k − 7 = − 12k = − 5
Find the value of c to complete the square:
r2 − 8r + c
  • r2 − 8r + ( [b/2] )2
  • r2 − 8r + ( [( − 8)/2] )2
  • r2 − 8r + 16
  • Factor:
    ( r + [(( − 8 ))/2] )2
( r − 4 )2
Find the value of c to complete the square:
g2 − 5g + c
  • g2 − 5g + ( [b/2] )2
  • g2 − 5g + [ [(( − 5 ))/2] ]2
  • g2 − 5g + [25/4]
  • Factor:
    ( g + [(( − 5 ))/2] )2
( g − [25/4] )2
Find the value of c to complete the square:
m2 + 10m + c
  • m2 + 10m + ( [b/2] )2
  • m2 + 10m + [(102)/4]
  • m2 + 10m + 25
  • Factor:
    ( m + [6/2] )2
  • ( m + [10/2] )2
( m + 5 )2
Solve by completing the square:
n2 − 6n − 9 = 11
  • n2 − 6n = 20
  • n2 − 6n + ( [b/2] )2 = 20 + ( [b/2] )2
  • n2 − 6n + [6/4]2 = 20 + [6/4]2
  • n2 − 6n + 9 = 20 + 9
  • n2 − 6n + 9 = 29
  • ( n + [b/2] )2 = 29
  • ( n + [(( − 6 ))/2] )2 = 29
  • ( n − 3 )2 = 29
  • n − 3 = ±√{29}
n = 3 ±√{29}
Solve by completing the square:
3x2 + 9x + 4 = 16
  • 3x2 + 9x = 12
  • x2 + 3x = 4
  • x2 + 3x + ( [b/2] )2 = 4 + ( [b/2] )2
  • x2 + 3x + [9/4] = 4 + [9/4]
  • x2 + 3x + [9/4] = 6[1/4]
  • x2 + [3/2] = ±√{6[1/4]}
x = − [3/2] ±√{6[1/4]}
Solve by completing the square:
v2 − 12v + 3 = 39
  • v2 − 12v = 36
  • v2 − 12v + [(( 12 )2)/4] = 36 + [(( 12 )2)/4]
  • v2 − 12v + 36 = 72
  • ( v + [( − 12)/2] )2 = 72
  • ( v − 6 )2 = 72
v − 6 = ±√{72} v = 6 ±√{72}
Solve by completing the square:
7w2 − 56w + 16 = 37
  • 7w2 − 56w = 21
  • w2 − 8w = 3
  • w2 − 8w + [(82)/4] = 3 + [(82)/4]
  • w2 − 8w + 16 = 19
  • ( w − 4 )2 = 19
  • w − 4 = ±√{19}
w = 4 ±√{19}
Solve using the quadratic formula:
5j2 + 6j + 1 = 0
  • j = [( − 6 ±√{62 − 4( 5 )( 1 )} )/2(5)]
  • j = [( − 6 ±√{16} )/10]
j = − [1/5], − 1
Solve using the quadratic formula:
8a2 − 11a + 3 = 0
  • a = [( − ( 11 ) ±√{( 11 )2 − 4( 8 )( 3 )} )/2( 8 )]
  • a = [( − ( 11 ) ±√{25} )/16]
  • a = [(11 ±5)/16]
a = 1,[3/8]
Solve using the quadratic formula:
b2 + 12b − 30 = 0
  • b = [( − 12 ±√{122 − 4( 1 )( 30 )} )/2(1)]
b = [( − 12 ±√{24} )/2]
Solve using the quadratic formula:
9r2 = 14r − 3
  • 9r2 − 14r + 3 = 0
  • r = [( − ( − 14 ) ±√{( − 14 )2 − 4( 9 )( 3 )} )/2( 9 )]
r = [(14 ±√{88} )/18]
Solve using the quadratic formula:
4s2 = − 7s − 2
  • 4s2 + 7s + 2 = 0
  • s = [( − 7 ±√{72 − 4( 4 )( 2 )} )/2( 4 )]
s = [( − 7 ±√{17} )/8]
Solve using the quadratic formula:
20m2 = 25m − 5
  • 4m2 = 5m − 1
  • 4m2 − 5m + 1 = 0
  • m = [( − ( − 1 ) ±√{( − 5 )2} − 4( 4 )( 1 ))/2(4)]
  • m = [(1 ±√9 )/8]
  • m = [(1 ±3)/8]
m = [1/2],[1/4]
Solve using the quadratic formula:
14y2 = 49y + 21
  • 2y2 = 7y + 3
  • 2y2 − 7y − 3 = 0
  • y = [( − ( − 7 ) ±√{( − 7 )2 − 4( 2 )( − 3 )} )/2( 2 )]
y = [(7 ±√{73} )/4]
Solve using the quadratic formula:
54n2 = − 81n + 18
  • 6n2 = 9n + 2
  • 6n + 9n − 2 = 0
  • n = [( − 9 ±√{92 − 4( 6 )( − 2 )} )/2( 6 )]
n = [( − 9 ±√{129} )/12]
Determine the number of real roots of 11x2 + 6x + 7 = 0D = 62 − 4( 11 )( 7 )
  • D = 36 − 308
  • D = − 272
No real solutions
b2 + 12b − 30 = 0
  • b = [( − 12 ±√{122 − 4( 1 )( 30 )} )/2( 1 )]
b = [( − 12 ±√{24} )/2]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Solving Quadratic Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Objectives 0:12
  • Solving Quadratic Equations 0:29
    • Linear Factors
    • Not All Quadratics Factor Easily
    • Principle of Square Roots
    • Completing the Square
    • Steps for Using Completing the Square
    • Completing the Square Works on All Quadratic Equations
    • The Quadratic Formula
    • Discriminants
    • Solving Quadratic Equations - Summary
  • Example 1 11:54
  • Example 2 13:03
  • Example 3 16:30
  • Example 4 21:29
  • Example 5 25:07

Transcription: Solving Quadratic Equations

Welcome back to

In this lesson we are going to continue on with our solving of equations by looking at some other solving techniques.0003

Some of these techniques that we have seen before is of course using factoring to solve quadratic equation.0014

What we are going to add to that is the square root principle, how to complete a square and everyone’s favorite, the quadratic formula.0019

One of the first ways that you can solve the quadratic equation is break it down into its linear factors.0032

When you do so, you can use the principle of 0 products.0039

We saw that the way that it works is we treat each of the factors as a number0042

and we know that one or possibly both of those factors must be equal to 0.0048

We borrowed each of them and set them equal to 0.0053

Then we solved each one individually and we got our solutions.0061

Now of course one downside to using a process like this is that it relies on our factoring techniques.0067

We have to be able to factor one of those quadratics and be able to grab them and set them equal to 0.0073

The reason why that is so much of a problem is because not all quadratics factor easily.0084

In fact, some of it will factor at all for searching around.0089

Let us take this one like 2x2 – 1 =3x.0093

Suppose we want to solve that quadratic equation, one of the first things we might do for trying to use that factoring technique0097

is we will go ahead and get everything over to one side.0104

Now that in itself is not too bad and then maybe we will attack this using something like reverse foil.0110

We only have one option to get 2x2, we must use a 2x and x.0117

There is only one way to get -1, we have to use 1 and -1.0123

The problem with this is if you go through and you check your outside and inside terms, here is what you get, -2x and 1x.0131

When those are combine it will give you -x, that is not the same as our middle term.0143

We know that factorization is not correct.0148

You might be saying to yourself we just made a small error.0152

Maybe this needs to be a -1 and the other one needs to be 1.0155

Well, let us try that one out and see if it turns out any better.0163

On the outside we have 2x, inside -1x, those combine to be 1x and unfortunately still not the same as our middle term.0168

There are no other possibilities for rearranging our 2 and 1, but those are the only possibilities that I could use my factoring process.0181

If I'm not able to factor this then how we are suppose to be able to find solutions.0191

After all, we still want to be able to find solutions for even quadratics like this.0197

For this reason alone, that we have to pick up some new methods for solving quadratics other than just the factoring process.0202

Let us first look at a nice one and see the method we can use on that type of quadratic.0210

In this type of quadratic, we only have one copy of x that is in the quadratic equation.0219

You do not have another x, just one copy of x2.0225

In these ones, the good news is you can solve directly.0229

This uses the principle of square roots.0233

What that principal says, is that if you have something like an x2 equals to a number then you can split this into 2.0237

Just put it into x equals the square root of that number and x equals the negative square root of that number.0247

What you are doing in practice is like you are taking the square root of both sides of your equation.0253

The reason why you are getting one being positive and one being negative is the square here erases information of what the sign used to be.0260

Then we will go ahead and give the ± to the other side.0274

x could have equal to 8, but maybe it was equal to -8.0277

It so this was a kind of nice if you can recognize that there is only a single copy of x running around and it is being squared.0283

For more complicated polynomials we will have to use a very special technique.0291

This one is known as completing the square.0296

What completing square does is it puts it into a new form and we can simply apply the principles of square roots to that new form.0299

The process of completing the square is a little bit lengthy, so let me walk you through how the completing the square works.0307

The first thing that you want to do is isolate your variables to one side so get all your x’s on one side.0315

If it does not have a variable, go ahead and put it on the other side of the equation for now.0321

You want to make sure that the coefficient on your x2 term is 1.0327

If it is anything other than 1 then go ahead and divide everything by that number.0331

We will be looking for a number that we can add to both sides of the equation so that it will go ahead and factor nicely.0338

There is a great technique for finding this number.0345

You take the coefficient on your x term, you divided it by 2 and you square it.0348

After you have added that number, then you can simply factor.0356

In fact, it should factor nicely at that point.0359

It should factor nicely.0363

If it does not factor nicely then double check your steps up to that one to make sure you follow them correctly.0369

We have gotten to that point, now we can use the principles of square roots.0377

This will be the square root of both sides and make sure that one of them is positive and one of them is negative.0380

Then we will solve directly for our variable.0388

This is quite a lengthy process using complete the square, but if you methodically go through all of the steps, it does get the job done.0391

The reason why I like completing the square so much is because it will work on any quadratic equation.0403

We do not have to worry if it is a special form, or if it is easily factorable, we can just use completing the square and it will work.0410

Of course there is one major downside to using a completing a square.0417

There are lots of necessary steps and it makes it very easy to make a mistake when trying to follow it.0421

The good news is all of these steps can be packaged up into a handy formula that we do not have quite as many steps to memorize.0428

This formula is known as the quadratic formula.0436

That is definitely something you want to know to be able to tackle many more difficult types of quadratics.0440

What the quadratic formula says is that if you are looking at a quadratic equation of the form ax2 + bx + c = 0, its solutions are given by this.0450

x = b ±√(b2 )- 4ac ÷2a.0461

It is quite a lot to keep track of but this is probably good one to memorize.0469

There is some good news, even though it is not the prettiest looking formula, it is built directly from completing the square.0478

Completing the square works on any quadratics so you may also use this on any quadratic equation.0485

It makes it extremely powerful for solving a lot of those quadratic equations.0491

Now that we have some more tools under our belt, let us take a look at using some of these new techniques.0498

To determine what type of solutions that these many techniques will produce, we can use what is called the discriminate to help us out.0511

The discriminate is actually just a part of the quadratic formula.0520

It is the part that sits underneath the square root.0525

The reason why that will tell us what type of solutions we get0528

is because depending on what type of numbers underneath the square root it could be real or it could be imaginary.0533

It could be rational or irrational.0538

Let us see how we can break that down.0541

Before looking at the part underneath the square root, that is the discriminate and say it is equal to 0.0545

What that tells us is that we will get exactly one real number solution.0552

If we are looking at the discriminate and it happens to be greater than 0, think of some positive number,0558

then we know we will have two different real number solutions.0566

This one is special because you can take it a little bit farther.0570

If that number is not only positive, but it is also a square number, your solutions will be rational.0574

If it is not a square number then it will be irrational.0582

And we just have one more case to worry about.0588

What happens if the discriminate is negative?0590

In that case you will have imaginary number solutions.0593

Think of earlier when we are dealing with a negative sign underneath the square root, these happen to be complex conjugates of each other,0598

but I will explain more about these in another lesson.0606

With so many new techniques to solve quadratics, how do you know what to use when just faced with any quadratic?0613

After all, some work for only certain quadratics if they are factorable and some techniques will work on any quadratic.0621

Here are some good tips on when you should use these.0628

The factoring technique that we picked up at the very beginning is one of the fastest methods you could use.0632

This means if you are looking at a quadratic and you can see that it is factorable0639

then go ahead and use that method because you will save yourself a lot of time.0644

If you end up struggling with it for quite a bit then maybe another technique is what should be used.0648

You can use the principle of square roots if you only have a single copy of x2.0656

This is a nice method because it is a very straightforward method that does not involve factoring and formulas.0661

You just solve directly.0668

If you are essentially stuck for too long and you know those methods worked out then you could try completing the square.0672

The good news is, it should be able to find your solutions, even though it does have quite a bit of steps.0679

In practice we do not usually use that one since the quadratic formula is built from the completing a square0685

and it will also work for any quadratic equation.0690

The reason why I like the quadratic formula it is a little bit easier to work with0695

and then we simply have to plugged everything into the formula and use it.0700

Make sure you identify your A, B and C terms and know that this one is faster than completing the square.0705

Let us try some of these techniques and see how it does.0716

We want to solve this first one using the square root principle.0721

To figure out why we even use that principle, notice how there is only one x and it is being squared.0726

There is no need to factor this one, we will just go ahead and work on getting x all by itself.0732

Add 5 to both sides and then we will divide both sides by 2.0739

And then we will take the square root of both sides.0754

This is where that property comes into play because I have x2 equals a number.0758

I either have x = √5 or x = √(-5).0764

Let us get some more space in there.0772

Just solve for this one directly.0779

Let us try the method of completing the square with this one.0785

2x2 + 4x – 2.0788

There are lots of steps involved so carefully watch how I walk through this one.0791

The very first thing I want to do is just start sorting things out.0796

I will get everything associated with a variable on one side, if it does not have a variable I will move it to the other side.0800

I'm adding 2 to both sides I do not have to worry about that 2 just yet.0809

I want to check the leading coefficient to make sure that it is a 1.0814

Unfortunately it looks like this one was not 1.0819

To make it a 1 I’m going to divide everything by 2.0822

2x2 ÷ 2 = x2, 4x ÷ 2= 2x and 2 ÷ 2 =1.0831

Now comes the part where I need to add something to this quadratic, so that it factors nicely.0842

The way we find out what this number is, is we look at the term in front of x.0849

I have a 2.0853

We are going to take that coefficient, divided by 2 and squared.0855

It gets a lot of questions like, should you always divide by 2 and square it?0861

The answer is yes it is always divided by 2 and squared.0865

2 ÷ 2 is 1 and 12 is 1.0869

This number right over here is what I need to add to both sides of my equation.0875

There is all the original numbers and there is that +1 on both sides, just to keep things balanced.0883

The reason why we are doing that is because on the left side it will now factor nicely0891

You could use the reverse foil method on that left side and you will get x + 1 and x + 1.0897

We just get a 2.0905

If you do add the correct number to both sides, these factors will always be the same.0908

If they are not the same then go back through and check your work.0916

Make sure you did not make a mistake, but they will always end up being the same.0919

We are going to write this as x + 12.0923

That condenses things down and actually at this point this is where I can use my principle of square roots.0930

We are going to use it because now I can take the square root of both sides to get rid of my square.0941

The x + 1 = √2 , I will have x + 1= -√2.0948

I can solve each of these directly.0959

x = -1 + √2 and x= -1 -√2 .0963

That one does take a lot more work, especially going through the entire process of completing the square.0974

But notice how it did find our solutions.0980

It even found our solutions, despite the fact that they were irrational solutions.0983

Unfortunately completing the square can be a clunky process, and that is what I want to demonstrate with this next example.0993

2x2 – 1 = 3.0999

Watch out even though the numbers are little bit more difficult to work with,1002

I will still go through all of the same steps in order to get my solution.1005

The very first thing I want to do is just get all of my x’s to one side.1010

If it does not have an x, I will move it to the other.1014

2x2 - 3x we will add the 1 to the other side 1.1018

Now I need to check my x2 term to see if it is 1.1026

It looks like this one is a 2 so I need to divide everything by 2.1030

That will give us x2 – 3/3x = ½.1041

Notice how we now have a lot more fractions showing up.1049

We have another method wrong it is just what the numbers turns out to be.1053

Now that we have gotten to this point I need to figure out what number I should add to both sides of my equation1057

so that the left side will factor nicely.1063

It comes from this number right here.1066

We will take it, divided by 2 and take that result and square it.1069

Just because it is a fraction it does not mean we can not work with it,1078

is the same as multiplying by ½ , so I get -3/42 which is the same as 9/16.1082

9/16 is the number that I should add to both sides.1093

Looking good, now that we have gotten to this stage the left side over here should factor nicely.1114

You could use method of reverse foil to help you out.1127

Know that your first terms will be x and two things that multiply to give you 9 would be a 3.1132

And two things that will multiply to give you 16 would be 4.1140

Both of these must be negative since our middle term is negative.1143

Let us see what is going on the other side.1148

I have ½ and I have 9/16.1150

What I can get those together by finding a common denominator, 8/16 + 9/16.1154

At this stage we have x -3/4 all of that is being squared and then all of that is equal to 17/16.1163

Okay so completing a square is a lot of work for us.1175

It gets our x’s together on the same side.1178

We can solve directly using the principle of square roots and taking the square root of both sides.1183

x - 3/4 I have the √17/16 or could be negative.1191

I’m going to represent those by saying it could be the + or minus of that square root.1199

I have one of my properties that says I can split up my square root over the top and bottom of a fraction.1207

The √17 or √16 or x -3/4 = ±√17÷4.1214

We are getting close to our actual solution, there is only one last thing I need to do.1225

Let us go ahead and add 3/4 to both sides.1229

I have x = ¾ ±√17÷4.1241

Since these have the same denominator I will go ahead and just write them as one large fraction.1250

That was quite a bit of work but notice how our answer was not a very nice answer.1261

It happens to be another irrational number.1267

We have no hope of finding irrational solutions like that using simply factoring.1270

Completing the square is an important process for manipulating quadratics and still being able to find solutions.1274

I will admit though it is quite a lengthy process, so be careful that you follow all the steps correctly.1281

A much faster method than completing the square is our quadratic formula.1291

The good news is, it will also work on any type of quadratic.1297

You just have to be careful to pickup the proper values and put them in their spot.1302

To help out, I’m going to write down the quadratic formula up in the corner here.1305

x= - b ±√(b2 )- 4ac ÷2a.1309

That way we can reference it when we need to solve this one.1323

In order for this to work out I need it set equal to 0 first.1327

Let us get everything on one side.1331

I’m going to organize things from the largest power to the smallest power.1334

We will have to subtract that 7 over and now it is set equal to 0 just fine.1339

My A value would be 3, B value would be 2 and the C value would be -7.1347

I’m going to take all of those and put them into the formula.1354

x is equal to negative the value of B which I said was 2 + or minus the square root.1360

That is where we will put in B again 22 -4 × the value of A × the value of C.1371

A and C all over 2 × the value of A.1384

You will see I just substituted the proper values into all of their spots.1396

We have to go through and very carefully simplify this down.1403

We have -2 ± 22 = 4.1409

I have that -4 × 3 × -7 I have to put all of those together.1417

What will that give me? I have 84 ÷ 6.1425

- 2 ±√88÷6.1436

It is important to go ahead and try and simplify this as much as possible.1445

I have √88 in there if you look at that as 4 × 22 and be able to take care of √4 and √2 separately.1453

√4 would be 2 and √22 has to stay and then I can cancel out 2 from the top and the bottom.1469


You will notice how this answer was not so nice but the quadratic formula helped us find it.1490

You just have to be very careful that you put it in the right spot and simplify carefully.1495

One common mistake is watching your signs for the part underneath the square root.1500

One last thing to do is to use our discriminate to determine the types of solutions we will get.1509

No matter what method we use by the way.1515

The discriminate is just that part underneath the square root.1518

Think of b2 – 4 × ac that is our discriminate.1522

We can use this even though the numbers might get a little large.1528

Here is my a, b, and c.1532

Let us just evaluate what this discriminate would be.1537

I have b2, just 70, -4 × a × c.1541

If I looked at 702, 7 × 7 is 49.1558

I have some 0s in there and then 4 × 25 would be 100.1566

100 × 49 = 49.1571

I'm getting for this first discriminate is that it is all equal to 0.1576

What is does that mean in terms of my solutions?1582

If I was to look at the rest of the quadratic formula in there it is telling me that this entire part is going to 0.1584

Either be adding 0 or subtracting 0.1592

I will have one real solution.1594

Let us try out the next one.1605

I have a =1, b=4, c=2.1609

B will be 4, -4a and c and they we will multiply all that together.1623

42 is 16, 4 × 2 is 8, 16 - 8 is 8.1633

This number is greater than 0 but it is not a square number.1645

What I can say is that I will have two real solutions.1659

Since it is not a square number they will be irrational.1667

Let us do one more.1677

Identify a, b, and c and we will put that into our discriminate.1680

b2 - 4 × a × c.1692

It is time to crunch it all down, -102 is 100 and I have -4 × 3 × 15.1697

I got 12 × 15, 2 × 1, 2 + 1.1707

I have 100 - 180 or - 80.1722

In this one our discriminant is less than 0.1729

That is an indication that we will have two imaginary solutions.1733

By using the discriminant you get a good idea of the solutions that you will get in your final answer.1747

You can use a variety of techniques to go ahead and solve those quadratics and get your final answer.1752

Know that factoring is usually the fastest, but quadratic formula will work on any type of quadratics.1758

Thank you for watching