Find three consecutive odd integers such that the square of the second decreased by four times the first is seven more than six times the third.

x, x+2, x+4

(x+2)^2 - 4(x) = 6(x + 4) + 7

x^2 + 4x + 4 - 4x = 6x + 24 + 7

x^2 - 6x - 27 = 0

x = 9 (∨ x = -3)

The three numbers are

9, 11, 13

but it could be also

-3, -1, 1