G(3y)=3y^2

Solution for G(3y)=3y^2 equation:

(3G)=3G^2

We move all terms to the left:

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

determiningTheFunctionDomain
-3G^2+3G=0

a = -3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-3)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-3}=\frac{-6}{-6}
=1
$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-3}=\frac{0}{-6}
=0
$