4(3c-8)+20=2(5c+4)+3c

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Solution for 4(3c-8)+20=2(5c+4)+3c equation: 



4(3c-8)+20=2(5c+4)+3c

We move all terms to the left:

4(3c-8)+20-(2(5c+4)+3c)=0

We multiply parentheses

12c-(2(5c+4)+3c)-32+20=0



We calculate terms in parentheses: -(2(5c+4)+3c), so:

2(5c+4)+3c

We add all the numbers together, and all the variables

3c+2(5c+4)

We multiply parentheses

3c+10c+8

We add all the numbers together, and all the variables

13c+8

Back to the equation:

-(13c+8)



We add all the numbers together, and all the variables

12c-(13c+8)-12=0

We get rid of parentheses

12c-13c-8-12=0

We add all the numbers together, and all the variables

-1c-20=0

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

-c=20

c=20/-1

c=-20

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