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

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

(^3-2+4)D=0

We multiply parentheses

D^2-2D+4D=0

We add all the numbers together, and all the variables

D^2+2D=0

a = 1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·1·0
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*1}=\frac{-4}{2}
=-2
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*1}=\frac{0}{2}
=0
$