(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^2-13i-5=0

a = 6; b = -13; c = -5;
Δ = b2-4ac
Δ = -132-4·6·(-5)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-17}{2*6}=\frac{-4}{12}
=-1/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+17}{2*6}=\frac{30}{12}
=2+1/2
$