F(2a)=5(2a)^2+6(2a)-6

Solution for F(2a)=5(2a)^2+6(2a)-6 equation:

(2F)=5(2F)^2+6(2F)-6

We move all terms to the left:

(2F)-(5(2F)^2+6(2F)-6)=0

We get rid of parentheses

-52F^2+2F-62F+6=0

We add all the numbers together, and all the variables

-52F^2-60F+6=0

a = -52; b = -60; c = +6;
Δ = b2-4ac
Δ = -602-4·(-52)·6
Δ = 4848
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:

$\sqrt{\Delta}=\sqrt{4848}=\sqrt{16*303}=\sqrt{16}*\sqrt{303}=4\sqrt{303}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-60)-4\sqrt{303}}{2*-52}=\frac{60-4\sqrt{303}}{-104}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-60)+4\sqrt{303}}{2*-52}=\frac{60+4\sqrt{303}}{-104}
$