-5(c-2)=20-5c+10-5(c-2)=20-5c+10

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Solution for -5(c-2)=20-5c+10-5(c-2)=20-5c+10 equation: 



-5(c-2)=20-5c+10-5(c-2)=20-5c+10

We move all terms to the left:

-5(c-2)-(20-5c+10-5(c-2))=0

We multiply parentheses

-5c-(20-5c+10-5(c-2))+10=0



We calculate terms in parentheses: -(20-5c+10-5(c-2)), so:

20-5c+10-5(c-2)

determiningTheFunctionDomain
-5c-5(c-2)+20+10

We add all the numbers together, and all the variables

-5c-5(c-2)+30

We multiply parentheses

-5c-5c+10+30

We add all the numbers together, and all the variables

-10c+40

Back to the equation:

-(-10c+40)



We get rid of parentheses

-5c+10c-40+10=0

We add all the numbers together, and all the variables

5c-30=0

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

5c=30

c=30/5

c=6

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