(5-2i)(1-3i)=0

Solution for (5-2i)(1-3i)=0 equation:

(5-2i)(1-3i)=0

We add all the numbers together, and all the variables

(-2i+5)(-3i+1)=0

We multiply parentheses ..

(+6i^2-2i-15i+5)=0

We get rid of parentheses

6i^2-2i-15i+5=0

We add all the numbers together, and all the variables

6i^2-17i+5=0

a = 6; b = -17; c = +5;
Δ = b2-4ac
Δ = -172-4·6·5
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-13}{2*6}=\frac{4}{12}
=1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+13}{2*6}=\frac{30}{12}
=2+1/2
$