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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

-4i^2-2i+4i+2=0

We add all the numbers together, and all the variables

-4i^2+2i+2=0

a = -4; b = 2; c = +2;
Δ = b2-4ac
Δ = 22-4·(-4)·2
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{-(2)-6}{2*-4}=\frac{-8}{-8}
=1
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-4}=\frac{4}{-8}
=-1/2
$