28-(2c+3)=(c+5)+c

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Solution for 28-(2c+3)=(c+5)+c equation: 



28-(2c+3)=(c+5)+c

We move all terms to the left:

28-(2c+3)-((c+5)+c)=0

We get rid of parentheses

-2c-((c+5)+c)-3+28=0



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

(c+5)+c

We add all the numbers together, and all the variables

c+(c+5)

We get rid of parentheses

c+c+5

We add all the numbers together, and all the variables

2c+5

Back to the equation:

-(2c+5)



We add all the numbers together, and all the variables

-2c-(2c+5)+25=0

We get rid of parentheses

-2c-2c-5+25=0

We add all the numbers together, and all the variables

-4c+20=0

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

-4c=-20

c=-20/-4

c=+5

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