10i-2(3i-8)=8-(i+5)

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Solution for 10i-2(3i-8)=8-(i+5) equation: 



10i-2(3i-8)=8-(i+5)

We move all terms to the left:

10i-2(3i-8)-(8-(i+5))=0

We multiply parentheses

10i-6i-(8-(i+5))+16=0



We calculate terms in parentheses: -(8-(i+5)), so:

8-(i+5)

determiningTheFunctionDomain
-(i+5)+8

We get rid of parentheses

-i-5+8

We add all the numbers together, and all the variables

-1i+3

Back to the equation:

-(-1i+3)



We add all the numbers together, and all the variables

4i-(-1i+3)+16=0

We get rid of parentheses

4i+1i-3+16=0

We add all the numbers together, and all the variables

5i+13=0

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

5i=-13

i=-13/5

i=-2+3/5

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