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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4F^2+F+4=0

a = -4; b = 1; c = +4;
Δ = b2-4ac
Δ = 12-4·(-4)·4
Δ = 65
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{65}}{2*-4}=\frac{-1-\sqrt{65}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{65}}{2*-4}=\frac{-1+\sqrt{65}}{-8}
$