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

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

(2F)=3F^2-4

We move all terms to the left:

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

We get rid of parentheses

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

a = -3; b = 2; c = +4;
Δ = b2-4ac
Δ = 22-4·(-3)·4
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{13}}{2*-3}=\frac{-2-2\sqrt{13}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*-3}=\frac{-2+2\sqrt{13}}{-6}
$