(4y+2)5y=0

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

(4y+2)5y=0

We multiply parentheses

20y^2+10y=0

a = 20; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·20·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*20}=\frac{-20}{40}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*20}=\frac{0}{40}
=0
$