3y(5+y)=0

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

3y(5+y)=0

We add all the numbers together, and all the variables

3y(y+5)=0

We multiply parentheses

3y^2+15y=0

a = 3; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·3·0
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*3}=\frac{-30}{6}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*3}=\frac{0}{6}
=0
$