(D^3-D^2-2D)y=8

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



(^3-^2-2)D=8
We move all terms to the left:
(^3-^2-2)D-(8)=0
We multiply parentheses
D^2+D^2-2D-8=0
We add all the numbers together, and all the variables
2D^2-2D-8=0
a = 2; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·2·(-8)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{17}}{2*2}=\frac{2-2\sqrt{17}}{4} $
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{17}}{2*2}=\frac{2+2\sqrt{17}}{4} $

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