G(3n)=2(3n)^2-3

Solution for G(3n)=2(3n)^2-3 equation:

(3G)=2(3G)^2-3

We move all terms to the left:

(3G)-(2(3G)^2-3)=0

We get rid of parentheses

-23G^2+3G+3=0

a = -23; b = 3; c = +3;
Δ = b2-4ac
Δ = 32-4·(-23)·3
Δ = 285
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{285}}{2*-23}=\frac{-3-\sqrt{285}}{-46}
$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{285}}{2*-23}=\frac{-3+\sqrt{285}}{-46}
$