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+3=0

a = -1; b = -3; c = +3;
Δ = b2-4ac
Δ = -32-4·(-1)·3
Δ = 21
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{21}}{2*-1}=\frac{3-\sqrt{21}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{21}}{2*-1}=\frac{3+\sqrt{21}}{-2}
$