(D^2+4D+2)Y=X^2-3x

Solution for (D^2+4D+2)Y=X^2-3x equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(D^2-3D)=0

We get rid of parentheses

-D^2+3D=0

We add all the numbers together, and all the variables

-1D^2+3D=0

a = -1; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-1)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-1}=\frac{-6}{-2}
=+3
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-1}=\frac{0}{-2}
=0
$