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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -(2(c+4)+c), so:

2(c+4)+c

We add all the numbers together, and all the variables

c+2(c+4)

We multiply parentheses

c+2c+8

We add all the numbers together, and all the variables

3c+8

Back to the equation:

-(3c+8)



We add all the numbers together, and all the variables

-3c-(3c+8)+24=0

We get rid of parentheses

-3c-3c-8+24=0

We add all the numbers together, and all the variables

-6c+16=0

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

-6c=-16

c=-16/-6

c=2+2/3

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