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

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

(^2+-2)D=2

We move all terms to the left:

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

We use the square of the difference formula

(^2-2)D-2=0

We multiply parentheses

D^2-2D-2=0

a = 1; b = -2; c = -2;
Δ = b2-4ac
Δ = -22-4·1·(-2)
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{3}}{2*1}=\frac{2-2\sqrt{3}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{3}}{2*1}=\frac{2+2\sqrt{3}}{2}
$