(D+2)(D^2-4D+3)y=0

Solution for (D+2)(D^2-4D+3)y=0 equation:

(+2)(^2-4+3)D=0

We add all the numbers together, and all the variables

2(^2-4+3)D=0

We multiply parentheses

2D^2-8D+6D=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
$