2y(3y-4)=0

Solution for 2y(3y-4)=0 equation:

2y(3y-4)=0

We multiply parentheses

6y^2-8y=0

a = 6; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·6·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*6}=\frac{0}{12}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*6}=\frac{16}{12}
=1+1/3
$