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

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



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

We multiply parentheses

D^2+D^2+4D-4D=0

We add all the numbers together, and all the variables

2D^2=0

a = 2; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·2·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$D=\frac{-b}{2a}=\frac{0}{4}=0$

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