y(3y-5)=0

Solution for y(3y-5)=0 equation:

y(3y-5)=0

We multiply parentheses

3y^2-5y=0

a = 3; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·3·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*3}=\frac{0}{6}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*3}=\frac{10}{6}
=1+2/3
$