3c+2(c+4)=128

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Solution for 3c+2(c+4)=128 equation: 



3c+2(c+4)=128

We move all terms to the left:

3c+2(c+4)-(128)=0

We multiply parentheses

3c+2c+8-128=0

We add all the numbers together, and all the variables

5c-120=0

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

5c=120

c=120/5

c=24

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