58-(-2c+3)=4(c+5)+c

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Solution for 58-(-2c+3)=4(c+5)+c equation: 



58-(-2c+3)=4(c+5)+c

We move all terms to the left:

58-(-2c+3)-(4(c+5)+c)=0

We get rid of parentheses

2c-(4(c+5)+c)-3+58=0



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

4(c+5)+c

We add all the numbers together, and all the variables

c+4(c+5)

We multiply parentheses

c+4c+20

We add all the numbers together, and all the variables

5c+20

Back to the equation:

-(5c+20)



We add all the numbers together, and all the variables

2c-(5c+20)+55=0

We get rid of parentheses

2c-5c-20+55=0

We add all the numbers together, and all the variables

-3c+35=0

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

-3c=-35

c=-35/-3

c=11+2/3

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