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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -8; b = -6; c = +9;
Δ = b2-4ac
Δ = -62-4·(-8)·9
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-8}=\frac{-12}{-16}
=3/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-8}=\frac{24}{-16}
=-1+1/2
$