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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3i^2-6i-9i-18)=0

We get rid of parentheses

-3i^2-6i-9i-18=0

We add all the numbers together, and all the variables

-3i^2-15i-18=0

a = -3; b = -15; c = -18;
Δ = b2-4ac
Δ = -152-4·(-3)·(-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{-(-15)-3}{2*-3}=\frac{12}{-6}
=-2
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3}{2*-3}=\frac{18}{-6}
=-3
$