(-3+2i)(-6+5i)=0

Solution for (-3+2i)(-6+5i)=0 equation:

(-3+2i)(-6+5i)=0

We add all the numbers together, and all the variables

(2i-3)(5i-6)=0

We multiply parentheses ..

(+10i^2-12i-15i+18)=0

We get rid of parentheses

10i^2-12i-15i+18=0

We add all the numbers together, and all the variables

10i^2-27i+18=0

a = 10; b = -27; c = +18;
Δ = b2-4ac
Δ = -272-4·10·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-3}{2*10}=\frac{24}{20}
=1+1/5
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+3}{2*10}=\frac{30}{20}
=1+1/2
$