(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+i+12=0

a = -6; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-6)·12
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-17}{2*-6}=\frac{-18}{-12}
=1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+17}{2*-6}=\frac{16}{-12}
=-1+1/3
$