2n(n+4)+n(n-2)-15=-3n(-n-1)+2n+6

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Solution for 2n(n+4)+n(n-2)-15=-3n(-n-1)+2n+6 equation: 



2n(n+4)+n(n-2)-15=-3n(-n-1)+2n+6

We move all terms to the left:

2n(n+4)+n(n-2)-15-(-3n(-n-1)+2n+6)=0

We add all the numbers together, and all the variables

2n(n+4)+n(n-2)-(-3n(-1n-1)+2n+6)-15=0

We multiply parentheses

2n^2+n^2+8n-2n-(-3n(-1n-1)+2n+6)-15=0



We calculate terms in parentheses: -(-3n(-1n-1)+2n+6), so:

-3n(-1n-1)+2n+6

We add all the numbers together, and all the variables

2n-3n(-1n-1)+6

We multiply parentheses

3n^2+2n+3n+6

We add all the numbers together, and all the variables

3n^2+5n+6

Back to the equation:

-(3n^2+5n+6)



We add all the numbers together, and all the variables

3n^2+6n-(3n^2+5n+6)-15=0

We get rid of parentheses

3n^2-3n^2+6n-5n-6-15=0

We add all the numbers together, and all the variables

n-21=0

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

n=21

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