(D-3^D+2)y=0

Solution for (D-3^D+2)y=0 equation:

(-3^+2)D=0

We multiply parentheses

-3D^2+2D=0

a = -3; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-3)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-3}=\frac{-4}{-6}
=2/3
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-3}=\frac{0}{-6}
=0
$