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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

2(c+3)+c

We add all the numbers together, and all the variables

c+2(c+3)

We multiply parentheses

c+2c+6

We add all the numbers together, and all the variables

3c+6

Back to the equation:

-(3c+6)



We add all the numbers together, and all the variables

-2c-(3c+6)+24=0

We get rid of parentheses

-2c-3c-6+24=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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