(3-2i)(4+1i)=0

Solution for (3-2i)(4+1i)=0 equation:

(3-2i)(4+1i)=0

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-2i^2-8i+3i+12)=0

We get rid of parentheses

-2i^2-8i+3i+12=0

We add all the numbers together, and all the variables

-2i^2-5i+12=0

a = -2; b = -5; c = +12;
Δ = b2-4ac
Δ = -52-4·(-2)·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-2}=\frac{16}{-4}
=-4
$