9+6i/3i=2-3i

Solution for 9+6i/3i=2-3i equation:

9+6i/3i=2-3i

We move all terms to the left:

9+6i/3i-(2-3i)=0


Domain of the equation: 3i!=0

i!=0/3

i!=0

i∈R


We add all the numbers together, and all the variables

6i/3i-(-3i+2)+9=0

We get rid of parentheses

6i/3i+3i-2+9=0

We multiply all the terms by the denominator


6i+3i*3i-2*3i+9*3i=0

Wy multiply elements

9i^2+6i-6i+27i=0

We add all the numbers together, and all the variables

9i^2+27i=0

a = 9; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·9·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$i_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*9}=\frac{-54}{18}
=-3
$
$i_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*9}=\frac{0}{18}
=0
$