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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4F^2-1F+7=0

a = -4; b = -1; c = +7;
Δ = b2-4ac
Δ = -12-4·(-4)·7
Δ = 113
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{113}}{2*-4}=\frac{1-\sqrt{113}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{113}}{2*-4}=\frac{1+\sqrt{113}}{-8}
$