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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+6i^2+4i-15i-10)=0

We get rid of parentheses

6i^2+4i-15i-10=0

We add all the numbers together, and all the variables

6i^2-11i-10=0

a = 6; b = -11; c = -10;
Δ = b2-4ac
Δ = -112-4·6·(-10)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*6}=\frac{-8}{12}
=-2/3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*6}=\frac{30}{12}
=2+1/2
$