z2+-1(-2+3)*z+(3-(-2))=0

Solution for z2+-1(-2+3)*z+(3-(-2))=0 equation:

z2+-1(-2+3)*z+(3-(-2))=0

We add all the numbers together, and all the variables

z2-11*z++(3-(-2))=0

We add all the numbers together, and all the variables

z^2-11z=0

a = 1; b = -11; c = 0;
Δ = b2-4ac
Δ = -112-4·1·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-11}{2*1}=\frac{0}{2}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+11}{2*1}=\frac{22}{2}
=11
$