-2(n+6)+3(2n-5)=3(n-4)+10

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Solution for -2(n+6)+3(2n-5)=3(n-4)+10 equation: 



-2(n+6)+3(2n-5)=3(n-4)+10

We move all terms to the left:

-2(n+6)+3(2n-5)-(3(n-4)+10)=0

We multiply parentheses

-2n+6n-(3(n-4)+10)-12-15=0



We calculate terms in parentheses: -(3(n-4)+10), so:

3(n-4)+10

We multiply parentheses

3n-12+10

We add all the numbers together, and all the variables

3n-2

Back to the equation:

-(3n-2)



We add all the numbers together, and all the variables

4n-(3n-2)-27=0

We get rid of parentheses

4n-3n+2-27=0

We add all the numbers together, and all the variables

n-25=0

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

n=25

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