4i+(1+i)+(2+i)+(3+i)+(4+i)=682

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Solution for 4i+(1+i)+(2+i)+(3+i)+(4+i)=682 equation:



4i+(1+i)+(2+i)+(3+i)+(4+i)=682
We move all terms to the left:
4i+(1+i)+(2+i)+(3+i)+(4+i)-(682)=0
We add all the numbers together, and all the variables
4i+(i+1)+(i+2)+(i+3)+(i+4)-682=0
We get rid of parentheses
4i+i+i+i+i+1+2+3+4-682=0
We add all the numbers together, and all the variables
8i-672=0
We move all terms containing i to the left, all other terms to the right
8i=672
i=672/8
i=84

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