Let $a,b,$ and $c$ be real numbers such that 
\[a+b+c=2, \text{ and}\]
\[a^2+b^2+c^2=12\]
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
$a+b= 2-c$. Now, by [Cauchy-Schwarz](https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz), we have that $(a^2+b^2) \ge \frac{(2-c)^2}{2}$. Therefore, we have that $\frac{(2-c)^2}{2}+c^2 \le 12$. We then find the roots of $c$ that satisfy equality and find the difference of the roots. This gives the answer, $\boxed{\textbf{(D)} \ \frac{16}{3}}$.
Note: We arrived at $(a^2+b^2) \ge \frac{(2-c)^2}{2}$ as we consider the two sets of numbers $(a,b)$ and $(1,1)$. From [Cauchy-Schwarz](https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz), we have that $(a^2+b^2)(1^2+1^2) \ge (2\cdot 1 -c\cdot 1)^2$. $a+b = 2-c$ which is how we got the second part of the inequality.