24-(3c+4)=2(c+4)+6

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Solution for 24-(3c+4)=2(c+4)+6 equation: 



24-(3c+4)=2(c+4)+6

We move all terms to the left:

24-(3c+4)-(2(c+4)+6)=0

We get rid of parentheses

-3c-(2(c+4)+6)-4+24=0



We calculate terms in parentheses: -(2(c+4)+6), so:

2(c+4)+6

We multiply parentheses

2c+8+6

We add all the numbers together, and all the variables

2c+14

Back to the equation:

-(2c+14)



We add all the numbers together, and all the variables

-3c-(2c+14)+20=0

We get rid of parentheses

-3c-2c-14+20=0

We add all the numbers together, and all the variables

-5c+6=0

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

-5c=-6

c=-6/-5

c=1+1/5

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