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

Solution for (D^2-D^2+2D+1)z=0 equation:

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

We multiply parentheses

D^2+D^2+2D+D=0

We add all the numbers together, and all the variables

2D^2+3D=0

a = 2; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*2}=\frac{-6}{4}
=-1+1/2
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*2}=\frac{0}{4}
=0
$