1. ## find hypotenuse

hi,

i need to solve x^2-18x+53=0 to find length of hypotenuse. i am unsure how to do this so some help would be greatly appreciated!

would i do quadratic equation on it??

thanks

2. Originally Posted by andyboy179
hi,

i need to solve x^2-18x+53=0 to find length of hypotenuse. i am unsure how to do this so some help would be greatly appreciated!

would i do quadratic equation on it??

thanks
This

$x^2-18x+53=0$

is a quadratic equation. To solve it for x use the quadratic formula:

The quadratic equation $ax^2 + bx +c = 0$ has the solution:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plug in the values of a, b and c which you already know.

3. i can do that but i will give me two different answers for x however i only need one what do i do with both the answers?

4. Choose the positive answer, since the side length of a triangle can't be negative (up to obscure fields of study )

5. Originally Posted by andyboy179
i can do that but i will give me two different answers for x however i only need one what do i do with both the answers?
I don't know! Since both solutions are positive you have to refer to the text of the question. If there aren't any specifications you can use both values.