What is the maximum possible value for the sum of the squares of the roots of $x^4+ax^3+bx^2+cx+d$ where $a$, $b$, $c$, and $d$ are $2$, $0$, $1$, and $7$ in some order?
1. Let the roots of the polynomial \(x^4 + ax^3 + bx^2 + cx + d\) be \(p, q, r, s\).
2. We need to find the maximum possible value for the sum of the squares of the roots, \(p^2 + q^2 + r^2 + s^2\).
3. Using the identity for the sum of squares of the roots, we have:
   \[
   p^2 + q^2 + r^2 + s^2 = (p + q + r + s)^2 - 2(pq + pr + ps + qr + qs + rs)
   \]
4. By Vieta's formulas, the sum of the roots \(p + q + r + s\) is equal to \(-a\), and the sum of the product of the roots taken two at a time \(pq + pr + ps + qr + qs + rs\) is equal to \(b\).
5. Substituting these into the identity, we get:
   \[
   p^2 + q^2 + r^2 + s^2 = (-a)^2 - 2b = a^2 - 2b
   \]
6. To maximize \(p^2 + q^2 + r^2 + s^2\), we need to maximize \(a^2\) and minimize \(b\).
7. Given that \(a, b, c, d\) are \(2, 0, 1, 7\) in some order, the maximum value of \(a\) is \(7\) and the minimum value of \(b\) is \(0\).
8. Substituting these values, we get:
   \[
   p^2 + q^2 + r^2 + s^2 = 7^2 - 2 \cdot 0 = 49 - 0 = 49
   \]

The final answer is \(\boxed{49}\).