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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2-3F+6=0

a = -3; b = -3; c = +6;
Δ = b2-4ac
Δ = -32-4·(-3)·6
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*-3}=\frac{-6}{-6}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*-3}=\frac{12}{-6}
=-2
$