12(c+3)=4(c+12)+6c

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Solution for 12(c+3)=4(c+12)+6c equation: 



12(c+3)=4(c+12)+6c

We move all terms to the left:

12(c+3)-(4(c+12)+6c)=0

We multiply parentheses

12c-(4(c+12)+6c)+36=0



We calculate terms in parentheses: -(4(c+12)+6c), so:

4(c+12)+6c

We add all the numbers together, and all the variables

6c+4(c+12)

We multiply parentheses

6c+4c+48

We add all the numbers together, and all the variables

10c+48

Back to the equation:

-(10c+48)



We get rid of parentheses

12c-10c-48+36=0

We add all the numbers together, and all the variables

2c-12=0

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

2c=12

c=12/2

c=6

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