28-(3c+4)=2(9c+3)+c

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Solution for 28-(3c+4)=2(9c+3)+c equation: 



28-(3c+4)=2(9c+3)+c

We move all terms to the left:

28-(3c+4)-(2(9c+3)+c)=0

We get rid of parentheses

-3c-(2(9c+3)+c)-4+28=0



We calculate terms in parentheses: -(2(9c+3)+c), so:

2(9c+3)+c

We add all the numbers together, and all the variables

c+2(9c+3)

We multiply parentheses

c+18c+6

We add all the numbers together, and all the variables

19c+6

Back to the equation:

-(19c+6)



We add all the numbers together, and all the variables

-3c-(19c+6)+24=0

We get rid of parentheses

-3c-19c-6+24=0

We add all the numbers together, and all the variables

-22c+18=0

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

-22c=-18

c=-18/-22

c=9/11

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