(-1+2j)*(3-2j)=0

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

(-1+2j)(3-2j)=0

We add all the numbers together, and all the variables

(2j-1)(-2j+3)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

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

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