5y(4y+3)=0

Solution for 5y(4y+3)=0 equation:

5y(4y+3)=0

We multiply parentheses

20y^2+15y=0

a = 20; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·20·0
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*20}=\frac{-30}{40}
=-3/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*20}=\frac{0}{40}
=0
$