(n+5)(n-1)=(n+3)(n+2)

Solution for (n+5)(n-1)=(n+3)(n+2) equation:

(n+5)(n-1)=(n+3)(n+2)

We move all terms to the left:

(n+5)(n-1)-((n+3)(n+2))=0

We multiply parentheses ..

(+n^2-1n+5n-5)-((n+3)(n+2))=0


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

(n+3)(n+2)

We multiply parentheses ..

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

We get rid of parentheses

n^2+2n+3n+6

We add all the numbers together, and all the variables

n^2+5n+6

Back to the equation:

-(n^2+5n+6)


We get rid of parentheses

n^2-n^2-1n+5n-5n-5-6=0

We add all the numbers together, and all the variables

-1n-11=0

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

-n=11

n=11/-1

n=-11