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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = 2; c = +3;
Δ = b2-4ac
Δ = 22-4·(-4)·3
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*-4}=\frac{-2-2\sqrt{13}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{13}}{2*-4}=\frac{-2+2\sqrt{13}}{-8}
$