2c+3=2(6+3c)+6c

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



2c+3=2(6+3c)+6c

We move all terms to the left:

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

We add all the numbers together, and all the variables

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



We calculate terms in parentheses: -(2(3c+6)+6c), so:

2(3c+6)+6c

We add all the numbers together, and all the variables

6c+2(3c+6)

We multiply parentheses

6c+6c+12

We add all the numbers together, and all the variables

12c+12

Back to the equation:

-(12c+12)



We get rid of parentheses

2c-12c-12+3=0

We add all the numbers together, and all the variables

-10c-9=0

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

-10c=9

c=9/-10

c=-9/10

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