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

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

(3)=2F^2+6F

We move all terms to the left:

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

We get rid of parentheses

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

a = -2; b = -6; c = +3;
Δ = b2-4ac
Δ = -62-4·(-2)·3
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*-2}=\frac{6-2\sqrt{15}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*-2}=\frac{6+2\sqrt{15}}{-4}
$