29(-2c+4)=2(c+5)+c

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Solution for 29(-2c+4)=2(c+5)+c equation: 



29(-2c+4)=2(c+5)+c

We move all terms to the left:

29(-2c+4)-(2(c+5)+c)=0

We multiply parentheses

-58c-(2(c+5)+c)+116=0



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

2(c+5)+c

We add all the numbers together, and all the variables

c+2(c+5)

We multiply parentheses

c+2c+10

We add all the numbers together, and all the variables

3c+10

Back to the equation:

-(3c+10)



We get rid of parentheses

-58c-3c-10+116=0

We add all the numbers together, and all the variables

-61c+106=0

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

-61c=-106

c=-106/-61

c=1+45/61

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