(D^2+D-6)y=1

Solution for (D^2+D-6)y=1 equation:

(^2+-6)D=1

We move all terms to the left:

(^2+-6)D-(1)=0

We use the square of the difference formula

(^2-6)D-1=0

We multiply parentheses

D^2-6D-1=0

a = 1; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·1·(-1)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{10}}{2*1}=\frac{6-2\sqrt{10}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{10}}{2*1}=\frac{6+2\sqrt{10}}{2}
$