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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

2(c+3)+3

We multiply parentheses

2c+6+3

We add all the numbers together, and all the variables

2c+9

Back to the equation:

-(2c+9)



We add all the numbers together, and all the variables

-3c-(2c+9)+32=0

We get rid of parentheses

-3c-2c-9+32=0

We add all the numbers together, and all the variables

-5c+23=0

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

-5c=-23

c=-23/-5

c=4+3/5

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