6-2(3c-5)=4(3c+2)-3c

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Solution for 6-2(3c-5)=4(3c+2)-3c equation: 



6-2(3c-5)=4(3c+2)-3c

We move all terms to the left:

6-2(3c-5)-(4(3c+2)-3c)=0

We multiply parentheses

-6c-(4(3c+2)-3c)+10+6=0



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

4(3c+2)-3c

We add all the numbers together, and all the variables

-3c+4(3c+2)

We multiply parentheses

-3c+12c+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

-6c-(9c+8)+16=0

We get rid of parentheses

-6c-9c-8+16=0

We add all the numbers together, and all the variables

-15c+8=0

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

-15c=-8

c=-8/-15

c=8/15

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