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

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

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

We multiply parentheses ..

(+3i^2+9i+2i+6)=0

We get rid of parentheses

3i^2+9i+2i+6=0

We add all the numbers together, and all the variables

3i^2+11i+6=0

a = 3; b = 11; c = +6;
Δ = b2-4ac
Δ = 112-4·3·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*3}=\frac{-18}{6}
=-3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*3}=\frac{-4}{6}
=-2/3
$