5y(2y-3)=0

Solution for 5y(2y-3)=0 equation:

5y(2y-3)=0

We multiply parentheses

10y^2-15y=0

a = 10; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·10·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*10}=\frac{0}{20}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*10}=\frac{30}{20}
=1+1/2
$