Find two consecutive odd integers such that the sum of their squares is 290

Find two consecutive odd integers such that the sum of their squares is 290.

Let the first odd integer be n.
The next odd integer is n + 2

Square them both:
n^2
(n + 2)^2 = n^2 + 4n + 4 from our expansion calculator

The sum of the squares equals 290
n^2 + n^2 + 4n + 4 = 290

Group like terms:
2n^2 + 4n + 4 = 290

Enter this quadratic into our search engine, and we get:
n = 11, n = -13

Which means the two consecutive odd integer are:
11 and 11 + 2 = 13. (11, 13)
-13 and -13 + 2 = -11 (-13, -11)