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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2+5F+4=0

a = -3; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·(-3)·4
Δ = 73
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{-(5)-\sqrt{73}}{2*-3}=\frac{-5-\sqrt{73}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*-3}=\frac{-5+\sqrt{73}}{-6}
$