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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -1; b = 5; c = +5;
Δ = b2-4ac
Δ = 52-4·(-1)·5
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3\sqrt{5}}{2*-1}=\frac{-5-3\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3\sqrt{5}}{2*-1}=\frac{-5+3\sqrt{5}}{-2}
$