we have the next equation
[tex]5x^2-36x+36=0[/tex]
we will use the formula to find the values of x of a second-grade equation
[tex]x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
where
a=5
b=-36
c=36
we substitute the values in the formula
[tex]x_{1,2}=\frac{36\pm\sqrt[]{36^2-4(5)(36)}}{2(5)}=\frac{36\pm\sqrt[]{576}}{10}=\frac{36\pm24}{10}[/tex][tex]x_1=\frac{36+24}{10}=\frac{60}{10}=6[/tex][tex]x_2=\frac{36-24}{10}=\frac{6}{5}[/tex]
the values of x are
x=6 and x=6/5
