3(i-3)-7(i-3)=4(2i-3)-8(2i+3)

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Solution for 3(i-3)-7(i-3)=4(2i-3)-8(2i+3) equation: 



3(i-3)-7(i-3)=4(2i-3)-8(2i+3)

We move all terms to the left:

3(i-3)-7(i-3)-(4(2i-3)-8(2i+3))=0

We multiply parentheses

3i-7i-(4(2i-3)-8(2i+3))-9+21=0



We calculate terms in parentheses: -(4(2i-3)-8(2i+3)), so:

4(2i-3)-8(2i+3)

We multiply parentheses

8i-16i-12-24

We add all the numbers together, and all the variables

-8i-36

Back to the equation:

-(-8i-36)



We add all the numbers together, and all the variables

-4i-(-8i-36)+12=0

We get rid of parentheses

-4i+8i+36+12=0

We add all the numbers together, and all the variables

4i+48=0

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

4i=-48

i=-48/4

i=-12

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