5(c+4)+2(3c-1)=3(2c-3)+4(c+12)

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Solution for 5(c+4)+2(3c-1)=3(2c-3)+4(c+12) equation: 



5(c+4)+2(3c-1)=3(2c-3)+4(c+12)

We move all terms to the left:

5(c+4)+2(3c-1)-(3(2c-3)+4(c+12))=0

We multiply parentheses

5c+6c-(3(2c-3)+4(c+12))+20-2=0



We calculate terms in parentheses: -(3(2c-3)+4(c+12)), so:

3(2c-3)+4(c+12)

We multiply parentheses

6c+4c-9+48

We add all the numbers together, and all the variables

10c+39

Back to the equation:

-(10c+39)



We add all the numbers together, and all the variables

11c-(10c+39)+18=0

We get rid of parentheses

11c-10c-39+18=0

We add all the numbers together, and all the variables

c-21=0

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

c=21

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