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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



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

(c+3)+3

We get rid of parentheses

c+3+3

We add all the numbers together, and all the variables

c+6

Back to the equation:

-(c+6)



We add all the numbers together, and all the variables

-2c-(c+6)+16=0

We get rid of parentheses

-2c-c-6+16=0

We add all the numbers together, and all the variables

-3c+10=0

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

-3c=-10

c=-10/-3

c=3+1/3

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