32-(3c+4)=2(c+5)=c

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Solution for 32-(3c+4)=2(c+5)=c equation: 



32-(3c+4)=2(c+5)=c

We move all terms to the left:

32-(3c+4)-(2(c+5))=0

We get rid of parentheses

-3c-(2(c+5))-4+32=0



We calculate terms in parentheses: -(2(c+5)), so:

2(c+5)

We multiply parentheses

2c+10

Back to the equation:

-(2c+10)



We add all the numbers together, and all the variables

-3c-(2c+10)+28=0

We get rid of parentheses

-3c-2c-10+28=0

We add all the numbers together, and all the variables

-5c+18=0

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

-5c=-18

c=-18/-5

c=3+3/5

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