-4(r+3)-(6-2r)=10

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Solution for -4(r+3)-(6-2r)=10 equation: 



-4(r+3)-(6-2r)=10

We move all terms to the left:

-4(r+3)-(6-2r)-(10)=0

We add all the numbers together, and all the variables

-4(r+3)-(-2r+6)-10=0

We multiply parentheses

-4r-(-2r+6)-12-10=0

We get rid of parentheses

-4r+2r-6-12-10=0

We add all the numbers together, and all the variables

-2r-28=0

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

-2r=28

r=28/-2

r=-14

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