4y(5y+2)=0

Solution for 4y(5y+2)=0 equation:

4y(5y+2)=0

We multiply parentheses

20y^2+8y=0

a = 20; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·20·0
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*20}=\frac{-16}{40}
=-2/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*20}=\frac{0}{40}
=0
$