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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

8i^2-14i+3=0

a = 8; b = -14; c = +3;
Δ = b2-4ac
Δ = -142-4·8·3
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-10}{2*8}=\frac{4}{16}
=1/4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+10}{2*8}=\frac{24}{16}
=1+1/2
$