F(-4)=x^2+4x-5

Solution for F(-4)=x^2+4x-5 equation:

(-4)=F^2+4F-5

We move all terms to the left:

(-4)-(F^2+4F-5)=0

We add all the numbers together, and all the variables

-(F^2+4F-5)-4=0

We get rid of parentheses

-F^2-4F+5-4=0

We add all the numbers together, and all the variables

-1F^2-4F+1=0

a = -1; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·(-1)·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{5}}{2*-1}=\frac{4-2\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{5}}{2*-1}=\frac{4+2\sqrt{5}}{-2}
$