(3i)(5-4i)=0

Solution for (3i)(5-4i)=0 equation:

(3i)(5-4i)=0

We add all the numbers together, and all the variables

3i(-4i+5)=0

We multiply parentheses

-12i^2+15i=0

a = -12; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·(-12)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*-12}=\frac{-30}{-24}
=1+1/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*-12}=\frac{0}{-24}
=0
$