3(4c-7)=28+5c

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Solution for 3(4c-7)=28+5c equation: 



3(4c-7)=28+5c

We move all terms to the left:

3(4c-7)-(28+5c)=0

We add all the numbers together, and all the variables

3(4c-7)-(5c+28)=0

We multiply parentheses

12c-(5c+28)-21=0

We get rid of parentheses

12c-5c-28-21=0

We add all the numbers together, and all the variables

7c-49=0

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

7c=49

c=49/7

c=7

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