(3-4i)*(-3+2i)=0

Solution for (3-4i)*(-3+2i)=0 equation:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-8i^2+12i+6i-9)=0

We get rid of parentheses

-8i^2+12i+6i-9=0

We add all the numbers together, and all the variables

-8i^2+18i-9=0

a = -8; b = 18; c = -9;
Δ = b2-4ac
Δ = 182-4·(-8)·(-9)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6}{2*-8}=\frac{-24}{-16}
=1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6}{2*-8}=\frac{-12}{-16}
=3/4
$