28-(3c+4)=2c+12+c

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Solution for 28-(3c+4)=2c+12+c equation: 



28-(3c+4)=2c+12+c

We move all terms to the left:

28-(3c+4)-(2c+12+c)=0

We add all the numbers together, and all the variables

-(3c+4)-(3c+12)+28=0

We get rid of parentheses

-3c-3c-4-12+28=0

We add all the numbers together, and all the variables

-6c+12=0

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

-6c=-12

c=-12/-6

c=+2

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