jmc

Let $p(x)=a{x}^2+bx+c.$ Then $$\begin{array}{rl}7a+b\sqrt{7}+c& =22,\\ 11a+b\sqrt{11}+c& =30.\end{array}$$Since $a,$ $b,$ and $c$ are rational, the only way that the equations above can hold is if $b=0.$ Then $$\begin{array}{rl}7a+c& =22,\\ 11a+c& =30.\end{array}$$Solving this system, we find $a=2$ and $c=8.$

Then $p(x)=2x^2+8,$ so $p(\sqrt{17})=2\cdot 17+8=\boxed{42}.$