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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-2F+3=0

a = -1; b = -2; c = +3;
Δ = b2-4ac
Δ = -22-4·(-1)·3
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-1}=\frac{-2}{-2}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-1}=\frac{6}{-2}
=-3
$