4y(2+y)=0

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

4y(2+y)=0

We add all the numbers together, and all the variables

4y(y+2)=0

We multiply parentheses

4y^2+8y=0

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