C+3(1+2c)-4=6c+2(c+3)

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Solution for C+3(1+2c)-4=6c+2(c+3) equation: 



+3(1+2C)-4=6C+2(C+3)

We move all terms to the left:

+3(1+2C)-4-(6C+2(C+3))=0

We add all the numbers together, and all the variables

3(2C+1)-(6C+2(C+3))-4=0

We multiply parentheses

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



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

6C+2(C+3)

We multiply parentheses

6C+2C+6

We add all the numbers together, and all the variables

8C+6

Back to the equation:

-(8C+6)



We add all the numbers together, and all the variables

6C-(8C+6)-1=0

We get rid of parentheses

6C-8C-6-1=0

We add all the numbers together, and all the variables

-2C-7=0

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

-2C=7

C=7/-2

C=-3+1/2

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