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

Solution for F(0)=4x^2+4x-4 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -4; b = -4; c = +4;
Δ = b2-4ac
Δ = -42-4·(-4)·4
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{5}}{2*-4}=\frac{4-4\sqrt{5}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{5}}{2*-4}=\frac{4+4\sqrt{5}}{-8}
$