F(x+5)=x^2+4x-8

Solution for F(x+5)=x^2+4x-8 equation:

(F+5)=F^2+4F-8

We move all terms to the left:

(F+5)-(F^2+4F-8)=0

We get rid of parentheses

-F^2+F-4F+5+8=0

We add all the numbers together, and all the variables

-1F^2-3F+13=0

a = -1; b = -3; c = +13;
Δ = b2-4ac
Δ = -32-4·(-1)·13
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{61}}{2*-1}=\frac{3-\sqrt{61}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{61}}{2*-1}=\frac{3+\sqrt{61}}{-2}
$