F(4x)=3x^2+4x-4

Solution for F(4x)=3x^2+4x-4 equation:

(4F)=3F^2+4F-4

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2+4=0

a = -3; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-3)·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*-3}=\frac{0-4\sqrt{3}}{-6}
=-\frac{4\sqrt{3}}{-6}
=-\frac{2\sqrt{3}}{-3}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*-3}=\frac{0+4\sqrt{3}}{-6}
=\frac{4\sqrt{3}}{-6}
=\frac{2\sqrt{3}}{-3}
$