Question

I have a problem. I have a picture of it because I can't type it

Answer 1

we have the next equation

`5x^2-36x+36=0`

we will use the formula to find the values of x of a second-grade equation

`x_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

where

a=5

b=-36

c=36

we substitute the values in the formula

`x_(1,2)=\frac{36\pm\sqrt[]{36^2-4(5)(36)}}{2(5)}=\frac{36\pm\sqrt[]{576}}{10}=(36\pm24)/(10)`
`x_1=(36+24)/(10)=(60)/(10)=6`
`x_2=(36-24)/(10)=(6)/(5)`

the values of x are

x=6 and x=6/5