(12-5i)(-3-2i)=0

Solution for (12-5i)(-3-2i)=0 equation:

(12-5i)(-3-2i)=0

We add all the numbers together, and all the variables

(-5i+12)(-2i-3)=0

We multiply parentheses ..

(+10i^2+15i-24i-36)=0

We get rid of parentheses

10i^2+15i-24i-36=0

We add all the numbers together, and all the variables

10i^2-9i-36=0

a = 10; b = -9; c = -36;
Δ = b2-4ac
Δ = -92-4·10·(-36)
Δ = 1521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1521}=39$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-39}{2*10}=\frac{-30}{20}
=-1+1/2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+39}{2*10}=\frac{48}{20}
=2+2/5
$