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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-4)·8
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{33}}{2*-4}=\frac{2-2\sqrt{33}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{33}}{2*-4}=\frac{2+2\sqrt{33}}{-8}
$