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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-4F+7=0

a = -1; b = -4; c = +7;
Δ = b2-4ac
Δ = -42-4·(-1)·7
Δ = 44
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:

$\sqrt{\Delta}=\sqrt{44}=\sqrt{4*11}=\sqrt{4}*\sqrt{11}=2\sqrt{11}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{11}}{2*-1}=\frac{4-2\sqrt{11}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{11}}{2*-1}=\frac{4+2\sqrt{11}}{-2}
$