2y(3y+2)=0

Solution for 2y(3y+2)=0 equation:

2y(3y+2)=0

We multiply parentheses

6y^2+4y=0

a = 6; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·6·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*6}=\frac{-8}{12}
=-2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*6}=\frac{0}{12}
=0
$