(2-3i)(-2+i)=0

Solution for (2-3i)(-2+i)=0 equation:

(2-3i)(-2+i)=0

We add all the numbers together, and all the variables

(-3i+2)(i-2)=0

We multiply parentheses ..

(-3i^2+6i+2i-4)=0

We get rid of parentheses

-3i^2+6i+2i-4=0

We add all the numbers together, and all the variables

-3i^2+8i-4=0

a = -3; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·(-3)·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*-3}=\frac{-12}{-6}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*-3}=\frac{-4}{-6}
=2/3
$