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

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



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

We move all terms to the left:

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

We multiply parentheses

4c-(2(3c+4)+3c)+12+6=0



We calculate terms in parentheses: -(2(3c+4)+3c), so:

2(3c+4)+3c

We add all the numbers together, and all the variables

3c+2(3c+4)

We multiply parentheses

3c+6c+8

We add all the numbers together, and all the variables

9c+8

Back to the equation:

-(9c+8)



We add all the numbers together, and all the variables

4c-(9c+8)+18=0

We get rid of parentheses

4c-9c-8+18=0

We add all the numbers together, and all the variables

-5c+10=0

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

-5c=-10

c=-10/-5

c=+2

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