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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2-4F+6=0

a = -3; b = -4; c = +6;
Δ = b2-4ac
Δ = -42-4·(-3)·6
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{22}}{2*-3}=\frac{4-2\sqrt{22}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{22}}{2*-3}=\frac{4+2\sqrt{22}}{-6}
$