algebra
posted by zainab on .
The sum of the squares of two consecutive odd integers decreased by the product of the integers is the same as sixtyseven.

(n)^2 + (n+2)^2  n(n+2)=67
solve for n, then n+2
posted by zainab on .
The sum of the squares of two consecutive odd integers decreased by the product of the integers is the same as sixtyseven.
(n)^2 + (n+2)^2  n(n+2)=67
solve for n, then n+2