-(3n-2)-(2n-7)+3=-6(n-5)-(2n+2)+4

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Solution for -(3n-2)-(2n-7)+3=-6(n-5)-(2n+2)+4 equation: 



-(3n-2)-(2n-7)+3=-6(n-5)-(2n+2)+4

We move all terms to the left:

-(3n-2)-(2n-7)+3-(-6(n-5)-(2n+2)+4)=0

We get rid of parentheses

-3n-2n-(-6(n-5)-(2n+2)+4)+2+7+3=0



We calculate terms in parentheses: -(-6(n-5)-(2n+2)+4), so:

-6(n-5)-(2n+2)+4

We multiply parentheses

-6n-(2n+2)+30+4

We get rid of parentheses

-6n-2n-2+30+4

We add all the numbers together, and all the variables

-8n+32

Back to the equation:

-(-8n+32)



We add all the numbers together, and all the variables

-5n-(-8n+32)+12=0

We get rid of parentheses

-5n+8n-32+12=0

We add all the numbers together, and all the variables

3n-20=0

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

3n=20

n=20/3

n=6+2/3

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