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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

2(c+3)+2

We multiply parentheses

2c+6+2

We add all the numbers together, and all the variables

2c+8

Back to the equation:

-(2c+8)



We add all the numbers together, and all the variables

-3c-(2c+8)+24=0

We get rid of parentheses

-3c-2c-8+24=0

We add all the numbers together, and all the variables

-5c+16=0

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

-5c=-16

c=-16/-5

c=3+1/5

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