Once again, working with stable vector bundles on $\mathbb{P}^2$ I have run into a question that is really out of my area. (Thanks to everybody who helped out with my last question!)

Let $D>9$ be a rational number which is not a square, and consider the quadratic irrational

$$\xi = \frac{-3 + \sqrt{D}}{2},$$

(I'd be willing to force $D$ to be an integer, and even to assume $D \equiv 5 \pmod{8}$, but I don't think it matters). Let $r$ be the positive integer such that

$$(2r+1)^2 < D < (2r+3)^2.$$

Numerous examples with Mathematica suggest that the continued fraction expansion of $\xi$ takes the form

$$\xi = [r-1;\overline{a_1,\ldots,a_k}],$$ where the last term of the repeating part is $a_k = 2r+1$.

For my particular situtation, I'd be happy enough to know that the number $2r+1$ appears somewhere in the expansion.

As I know almost nothing about continued fractions aside from statements of the basic results, I haven't the slightest idea how to prove something like this. I also don't have a source which does much more advanced things than show that the expansion of a quadratic irrational always repeats. Are statements like this well-known? And where should I look for more advanced theory relevant to this problem?

In case it helps, the original form I came to this number is as follows. Put $$ q = \frac{1}{8}(D-5). $$ Then $\xi$ is a solution of the equation $$\frac{1}{2}(x^2+3x+1) = q.$$ Thanks!

EDIT: At request, here is what Mathematica gives for the continued fractions for some $D$:

$D=5: [0;-2,\overline{-1}]$ (but I am requiring $D>9$)

$D=10: [0;\overline{12,3}]$

$D=13: [0;\overline{3}]$

$D=141: [4;\overline{2,3,2,11}]$

(need more examples? Just ask!)