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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6i^2+17i-12=0

a = -6; b = 17; c = -12;
Δ = b2-4ac
Δ = 172-4·(-6)·(-12)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*-6}=\frac{-18}{-12}
=1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*-6}=\frac{-16}{-12}
=1+1/3
$