(3+2j)(-6+j)=0

Solution for (3+2j)(-6+j)=0 equation:

(3+2j)(-6+j)=0

We add all the numbers together, and all the variables

(2j+3)(j-6)=0

We multiply parentheses ..

(+2j^2-12j+3j-18)=0

We get rid of parentheses

2j^2-12j+3j-18=0

We add all the numbers together, and all the variables

2j^2-9j-18=0

a = 2; b = -9; c = -18;
Δ = b2-4ac
Δ = -92-4·2·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-15}{2*2}=\frac{-6}{4}
=-1+1/2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+15}{2*2}=\frac{24}{4}
=6
$