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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-12i^2+6i-16i+8)=0

We get rid of parentheses

-12i^2+6i-16i+8=0

We add all the numbers together, and all the variables

-12i^2-10i+8=0

a = -12; b = -10; c = +8;
Δ = b2-4ac
Δ = -102-4·(-12)·8
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-22}{2*-12}=\frac{-12}{-24}
=1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+22}{2*-12}=\frac{32}{-24}
=-1+1/3
$