3i-(7i+12)=2(i-3)

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Solution for 3i-(7i+12)=2(i-3) equation: 



3i-(7i+12)=2(i-3)

We move all terms to the left:

3i-(7i+12)-(2(i-3))=0

We get rid of parentheses

3i-7i-(2(i-3))-12=0



We calculate terms in parentheses: -(2(i-3)), so:

2(i-3)

We multiply parentheses

2i-6

Back to the equation:

-(2i-6)



We add all the numbers together, and all the variables

-4i-(2i-6)-12=0

We get rid of parentheses

-4i-2i+6-12=0

We add all the numbers together, and all the variables

-6i-6=0

We move all terms containing i to the left, all other terms to the right

-6i=6

i=6/-6

i=-1

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