<h3>
Answer: x = 6</h3>
======================================================
Work Shown:


Those are the possible solutions, but plugging x = -14 back into the original equation will lead to an error. So we rule x = -14 out
x = 6 works as a solution however
To find the hypotenuse of the triangle
you use the length of the legs
Plug it into the Pythagorean Theorem
so


is one of the legs
and

is the other leg
then

is the hypothenuse