(D^2-4D+13)y=0

Solution for (D^2-4D+13)y=0 equation:

(^2-4+13)D=0

We multiply parentheses

D^2-4D+13D=0

We add all the numbers together, and all the variables

D^2+9D=0

a = 1; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·1·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*1}=\frac{-18}{2}
=-9
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*1}=\frac{0}{2}
=0
$