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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

4(c+5)+5

We multiply parentheses

4c+20+5

We add all the numbers together, and all the variables

4c+25

Back to the equation:

-(4c+25)



We add all the numbers together, and all the variables

-2c-(4c+25)+55=0

We get rid of parentheses

-2c-4c-25+55=0

We add all the numbers together, and all the variables

-6c+30=0

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

-6c=-30

c=-30/-6

c=+5

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