Let $$p$$ and $$q$$ be two positive numbers such that $$p + q = 2$$ and $$p^4 + q^4 = 272$$. Then $$p$$ and $$q$$ are roots of the equation:

We are given $$p + q = 2$$ and $$p^4 + q^4 = 272$$, and we need to find the quadratic equation whose roots are $$p$$ and $$q$$.

We know that $$p^2 + q^2 = (p + q)^2 - 2pq = 4 - 2pq$$.

Now, $$p^4 + q^4 = (p^2 + q^2)^2 - 2p^2q^2$$. Substituting, we get $$(4 - 2pq)^2 - 2(pq)^2 = 272$$.

Let $$s = pq$$. Then $$16 - 16s + 4s^2 - 2s^2 = 272$$, which gives $$2s^2 - 16s + 16 = 272$$.

So $$2s^2 - 16s - 256 = 0$$, or $$s^2 - 8s - 128 = 0$$.

Using the quadratic formula, $$s = \frac{8 \pm \sqrt{64 + 512}}{2} = \frac{8 \pm \sqrt{576}}{2} = \frac{8 \pm 24}{2}$$.

This gives $$s = 16$$ or $$s = -8$$.

The quadratic equation with sum of roots $$p + q = 2$$ and product $$pq = 16$$ is $$x^2 - 2x + 16 = 0$$, and with $$pq = -8$$ it is $$x^2 - 2x - 8 = 0$$. Among the given options, only $$x^2 - 2x + 16 = 0$$ is present.

Hence, the correct answer is Option D.