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

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

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

We add all the numbers together, and all the variables

3i(4i-2)=0

We multiply parentheses

12i^2-6i=0

a = 12; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·12·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*12}=\frac{0}{24}
=0
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*12}=\frac{12}{24}
=1/2
$