(D^2-2D-3y)y=0

Solution for (D^2-2D-3y)y=0 equation:

(^2-2-3D)D=0

We multiply parentheses

D^2-3D^2-2D=0

We add all the numbers together, and all the variables

-2D^2-2D=0

a = -2; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-2)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-2}=\frac{0}{-4}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-2}=\frac{4}{-4}
=-1
$