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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-4)·4
Δ = 68
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:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*-4}=\frac{-2-2\sqrt{17}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*-4}=\frac{-2+2\sqrt{17}}{-8}
$