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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1F^2-4F+10=0

a = -1; b = -4; c = +10;
Δ = b2-4ac
Δ = -42-4·(-1)·10
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{14}}{2*-1}=\frac{4-2\sqrt{14}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{14}}{2*-1}=\frac{4+2\sqrt{14}}{-2}
$