I have calculated the continued fraction of $\alpha=\frac{6+\sqrt{47}}{11}$ which equals $\overline{[1,5,1,12]}$. Now I am asked to calculated the cont. fraction of $\sqrt{47}$ using this result. I am not sure whether there is a simple formula to calculate the continued fraction of $\sqrt{47}=11\alpha-6$.
I know the answer to be $\sqrt{47}=[6,\overline{1,5,1,12}]$ (checked by Mathematica) but it's not clear how to arrive at this result using our previous answer.