(D^4-D^2)y=2

Solution for (D^4-D^2)y=2 equation:

(^4-^2)D=2

We move all terms to the left:

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

We multiply parentheses

D^2+D^2-2=0

We add all the numbers together, and all the variables

2D^2-2=0

a = 2; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·2·(-2)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*2}=\frac{-4}{4}
=-1
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*2}=\frac{4}{4}
=1
$