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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = 4; c = +4;
Δ = b2-4ac
Δ = 42-4·(-3)·4
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-3}=\frac{-12}{-6}
=+2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-3}=\frac{4}{-6}
=-2/3
$