2y(4y-2)=0

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

2y(4y-2)=0

We multiply parentheses

8y^2-4y=0

a = 8; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·8·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*8}=\frac{0}{16}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*8}=\frac{8}{16}
=1/2
$