4y(3y-9)=0

Solution for 4y(3y-9)=0 equation:

4y(3y-9)=0

We multiply parentheses

12y^2-36y=0

a = 12; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·12·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*12}=\frac{0}{24}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*12}=\frac{72}{24}
=3
$