(D^3-14D+8)y=0

Solution for (D^3-14D+8)y=0 equation:

(^3-14+8)D=0

We multiply parentheses

D^2-14D+8D=0

We add all the numbers together, and all the variables

D^2-6D=0

a = 1; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*1}=\frac{0}{2}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*1}=\frac{12}{2}
=6
$