 Welcome Geometry students to another example of the Pythagorean theorem. Have your pencil and paper ready to follow along as this one has a lots of algebra in it and it will be a little bit tougher to solve. a squared plus b squared equals c squared is the Pythagorean theorem. So we're going to substitute in the legs x plus 3 squared and x plus 10 squared. And we're going to set that equal to that long square root of x squared plus 50x. I'm just going to put dot dot dot. Now remember you have to foil when you are multiplying out x plus 3 squared and x plus 10 squared. So x plus 3 squared is x squared plus 6x plus 9 and x plus 10 squared is x squared plus 20x plus 100. Now if we add like terms we're going to get 2x squared plus 26x plus 109 equals that c squared which was x squared plus 50x plus 65. And I'm sure you remember when you square a square root that whole square root symbol is going to go away. So we're going to be left with 2x squared plus 26x plus 109 equals x squared plus 50x plus 65. If we have x squared it's important to set everything equal to 0. So we're going to simplify this and bring everything over on the right side to the left side. So we're going to subtract x squared and get x squared. We're going to subtract 50x and get negative 24x and we're going to subtract 65 and get 44. And now we can unfoil it, factor it, and solve it. So our next step is going to be to try to find something that multiplies to 44 and adds to negative 24. So it's like foiling in reverse and you would use 22 and 2. You can check your answer by foiling. If you foiled out x minus 22 times x minus 2 you'd get x squared minus 24x plus 44. Okay, to finish the problem I'm going to slide everything up a little bit because we're running out of space and we're going to take our factors and we're going to set them equal to 0. So we're going to take x minus 22 and set it equal to 0 and get x equals 22 and we're going to set x minus 2 equal to 0 and get x equals to 2. Our last step should be to check those answers back into the original problem and see if 22 and 2 are actually okay to substitute in for x in all of these spots. And if you check it in you do get positive numbers. 22 into this leg would be 25, this would be, x plus 10 would be 32. So you just want to make sure none of the side lengths are negative and that's the solution to this problem. Kind of a wicked long algebra problem but kind of fun too with all the factoring involved. So the final answer would be x equals 22 or 2 and you can see that at the bottom.