(D^3-D^2-8D+1)y=0

Solution for (D^3-D^2-8D+1)y=0 equation:

(^3-^2-8+1)D=0

We multiply parentheses

D^2+D^2-8D+D=0

We add all the numbers together, and all the variables

2D^2-7D=0

a = 2; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·2·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*2}=\frac{0}{4}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*2}=\frac{14}{4}
=3+1/2
$