(D^2+3D+2)y=12X^2

Solution for (D^2+3D+2)y=12X^2 equation:

(^2+3+2)D=12D^2

We move all terms to the left:

(^2+3+2)D-(12D^2)=0

determiningTheFunctionDomain
-12D^2+(^2+3+2)D=0

We multiply parentheses

-12D^2+D^2+3D+2D=0

We add all the numbers together, and all the variables

-11D^2+5D=0

a = -11; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-11)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-11}=\frac{-10}{-22}
=5/11
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-11}=\frac{0}{-22}
=0
$