(D^2+2D+5)y=15

Solution for (D^2+2D+5)y=15 equation:

(^2+2+5)D=15

We move all terms to the left:

(^2+2+5)D-(15)=0

We multiply parentheses

D^2+2D+5D-15=0

We add all the numbers together, and all the variables

D^2+7D-15=0

a = 1; b = 7; c = -15;
Δ = b2-4ac
Δ = 72-4·1·(-15)
Δ = 109
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{109}}{2*1}=\frac{-7-\sqrt{109}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{109}}{2*1}=\frac{-7+\sqrt{109}}{2}
$