F=2(2a^2)+3(2a)-6

Solution for F=2(2a^2)+3(2a)-6 equation:

=2(2F^2)+3(2F)-6

We move all terms to the left:

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

We get rid of parentheses

-22F^2-32F+6=0

a = -22; b = -32; c = +6;
Δ = b2-4ac
Δ = -322-4·(-22)·6
Δ = 1552
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:

$\sqrt{\Delta}=\sqrt{1552}=\sqrt{16*97}=\sqrt{16}*\sqrt{97}=4\sqrt{97}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-4\sqrt{97}}{2*-22}=\frac{32-4\sqrt{97}}{-44}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+4\sqrt{97}}{2*-22}=\frac{32+4\sqrt{97}}{-44}
$