2i-3=2(6-i)+7i

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Solution for 2i-3=2(6-i)+7i equation: 



2i-3=2(6-i)+7i

We move all terms to the left:

2i-3-(2(6-i)+7i)=0

We add all the numbers together, and all the variables

2i-(2(-1i+6)+7i)-3=0



We calculate terms in parentheses: -(2(-1i+6)+7i), so:

2(-1i+6)+7i

We add all the numbers together, and all the variables

7i+2(-1i+6)

We multiply parentheses

7i-2i+12

We add all the numbers together, and all the variables

5i+12

Back to the equation:

-(5i+12)



We get rid of parentheses

2i-5i-12-3=0

We add all the numbers together, and all the variables

-3i-15=0

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

-3i=15

i=15/-3

i=-5

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