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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-6i^2+3i-8i+4)=0

We get rid of parentheses

-6i^2+3i-8i+4=0

We add all the numbers together, and all the variables

-6i^2-5i+4=0

a = -6; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·(-6)·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-11}{2*-6}=\frac{-6}{-12}
=1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+11}{2*-6}=\frac{16}{-12}
=-1+1/3
$