(4c+20)+(4c+20)+(3c-15)+(3c-15)=3600

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Solution for (4c+20)+(4c+20)+(3c-15)+(3c-15)=3600 equation: 



(4c+20)+(4c+20)+(3c-15)+(3c-15)=3600

We move all terms to the left:

(4c+20)+(4c+20)+(3c-15)+(3c-15)-(3600)=0

We get rid of parentheses

4c+4c+3c+3c+20+20-15-15-3600=0

We add all the numbers together, and all the variables

14c-3590=0

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

14c=3590

c=3590/14

c=256+3/7

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