56(3c+4)=4(c+5)+c

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Solution for 56(3c+4)=4(c+5)+c equation:



56(3c+4)=4(c+5)+c
We move all terms to the left:
56(3c+4)-(4(c+5)+c)=0
We multiply parentheses
168c-(4(c+5)+c)+224=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 get rid of parentheses
168c-5c-20+224=0
We add all the numbers together, and all the variables
163c+204=0
We move all terms containing c to the left, all other terms to the right
163c=-204
c=-204/163
c=-1+41/163

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