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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-1F+10=0

a = -1; b = -1; c = +10;
Δ = b2-4ac
Δ = -12-4·(-1)·10
Δ = 41
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{-(-1)-\sqrt{41}}{2*-1}=\frac{1-\sqrt{41}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{41}}{2*-1}=\frac{1+\sqrt{41}}{-2}
$