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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-15i^2-10i-9i-6)=0

We get rid of parentheses

-15i^2-10i-9i-6=0

We add all the numbers together, and all the variables

-15i^2-19i-6=0

a = -15; b = -19; c = -6;
Δ = b2-4ac
Δ = -192-4·(-15)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*-15}=\frac{18}{-30}
=-3/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*-15}=\frac{20}{-30}
=-2/3
$