F(-x)=2x^2+5x

Solution for F(-x)=2x^2+5x equation:

(-F)=2F^2+5F

We move all terms to the left:

(-F)-(2F^2+5F)=0

We add all the numbers together, and all the variables

(-1F)-(2F^2+5F)=0

We get rid of parentheses

-2F^2-1F-5F=0

We add all the numbers together, and all the variables

-2F^2-6F=0

a = -2; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·(-2)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*-2}=\frac{0}{-4}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*-2}=\frac{12}{-4}
=-3
$