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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4F^2-3F+4=0

a = -4; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-4)·4
Δ = 73
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{-(-3)-\sqrt{73}}{2*-4}=\frac{3-\sqrt{73}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{73}}{2*-4}=\frac{3+\sqrt{73}}{-8}
$