(242-c)/2+c+28=3c+242-c

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Solution for (242-c)/2+c+28=3c+242-c equation: 



(242-c)/2+c+28=3c+242-c

We move all terms to the left:

(242-c)/2+c+28-(3c+242-c)=0

We add all the numbers together, and all the variables

(-1c+242)/2+c-(2c+242)+28=0

We add all the numbers together, and all the variables

c+(-1c+242)/2-(2c+242)+28=0

We get rid of parentheses

c+(-1c+242)/2-2c-242+28=0

We multiply all the terms by the denominator


c*2+(-1c+242)-2c*2-242*2+28*2=0

We add all the numbers together, and all the variables

c*2+(-1c+242)-2c*2-428=0

Wy multiply elements

2c+(-1c+242)-4c-428=0

We get rid of parentheses

2c-1c-4c+242-428=0

We add all the numbers together, and all the variables

-3c-186=0

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

-3c=186

c=186/-3

c=-62

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