5y(3y-5)=0

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

5y(3y-5)=0

We multiply parentheses

15y^2-25y=0

a = 15; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·15·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*15}=\frac{0}{30}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*15}=\frac{50}{30}
=1+2/3
$