(1-9i)(3+2i)=0

Solution for (1-9i)(3+2i)=0 equation:

(1-9i)(3+2i)=0

We add all the numbers together, and all the variables

(-9i+1)(2i+3)=0

We multiply parentheses ..

(-18i^2-27i+2i+3)=0

We get rid of parentheses

-18i^2-27i+2i+3=0

We add all the numbers together, and all the variables

-18i^2-25i+3=0

a = -18; b = -25; c = +3;
Δ = b2-4ac
Δ = -252-4·(-18)·3
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-29}{2*-18}=\frac{-4}{-36}
=1/9
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+29}{2*-18}=\frac{54}{-36}
=-1+1/2
$