(8-3i)(4-2i)=0

Solution for (8-3i)(4-2i)=0 equation:

(8-3i)(4-2i)=0

We add all the numbers together, and all the variables

(-3i+8)(-2i+4)=0

We multiply parentheses ..

(+6i^2-12i-16i+32)=0

We get rid of parentheses

6i^2-12i-16i+32=0

We add all the numbers together, and all the variables

6i^2-28i+32=0

a = 6; b = -28; c = +32;
Δ = b2-4ac
Δ = -282-4·6·32
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*6}=\frac{24}{12}
=2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*6}=\frac{32}{12}
=2+2/3
$