(4+6i)(8-2i)=0

Solution for (4+6i)(8-2i)=0 equation:

(4+6i)(8-2i)=0

We add all the numbers together, and all the variables

(6i+4)(-2i+8)=0

We multiply parentheses ..

(-12i^2+48i-8i+32)=0

We get rid of parentheses

-12i^2+48i-8i+32=0

We add all the numbers together, and all the variables

-12i^2+40i+32=0

a = -12; b = 40; c = +32;
Δ = b2-4ac
Δ = 402-4·(-12)·32
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-56}{2*-12}=\frac{-96}{-24}
=+4
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+56}{2*-12}=\frac{16}{-24}
=-2/3
$