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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

F^2+5F+4=0

a = 1; b = 5; c = +4;
Δ = b2-4ac
Δ = 52-4·1·4
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-3}{2*1}=\frac{-8}{2}
=-4
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+3}{2*1}=\frac{-2}{2}
=-1
$