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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4i^2-24i+8i+48)=0

We get rid of parentheses

-4i^2-24i+8i+48=0

We add all the numbers together, and all the variables

-4i^2-16i+48=0

a = -4; b = -16; c = +48;
Δ = b2-4ac
Δ = -162-4·(-4)·48
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-32}{2*-4}=\frac{-16}{-8}
=+2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+32}{2*-4}=\frac{48}{-8}
=-6
$