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

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

(4F^2)=2F+8

We move all terms to the left:

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

We get rid of parentheses

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}
$