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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -((4c+20)), so:

(4c+20)

We get rid of parentheses

4c+20

Back to the equation:

-(4c+20)



We get rid of parentheses

3c-4c-20-15=0

We add all the numbers together, and all the variables

-1c-35=0

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

-c=35

c=35/-1

c=-35

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