The sum of the squares of 2 consecutive negative integers is 41. What are the numbers?
Which of the following equations is the result of using the factoring method to solve the problem?

(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0

Question

The sum of the squares of 2 consecutive negative integers is 41. What are the numbers?
Which of the following equations is the result of using the factoring method to solve the problem? 

(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0

anonymous · Answer

ok maybe try 
$$n^2+(n+1)^2=41$$ and see what happens

anonymous · Answer

I got (n+5)(n-4)

anonymous · Answer

multiplying out gives 
$$n^2+n^2+2n+1=41$$ subtract and get 
$$2n^2+2n-40=0$$ divide by 2 and get 
$$n^2+n-20=0$$ then 
$$(n+5)(n-4)=0$$

anonymous · Answer

yeah i got the same as you