20-(3c+4)=2(c+6)+c

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Solution for 20-(3c+4)=2(c+6)+c equation: 



20-(3c+4)=2(c+6)+c

We move all terms to the left:

20-(3c+4)-(2(c+6)+c)=0

We get rid of parentheses

-3c-(2(c+6)+c)-4+20=0



We calculate terms in parentheses: -(2(c+6)+c), so:

2(c+6)+c

We add all the numbers together, and all the variables

c+2(c+6)

We multiply parentheses

c+2c+12

We add all the numbers together, and all the variables

3c+12

Back to the equation:

-(3c+12)



We add all the numbers together, and all the variables

-3c-(3c+12)+16=0

We get rid of parentheses

-3c-3c-12+16=0

We add all the numbers together, and all the variables

-6c+4=0

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

-6c=-4

c=-4/-6

c=2/3

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