Continued Fraction Form of sqrt(6)

Question

I have to find the continued fraction form of sqrt(6). I have tried it, and have the answers but I can't get to the correct answer. If someone could help me that would be much appreciated. Thank you!

Answer 1

Hint: $2 < \sqrt{6} < 3$ so it starts $2 + 1/\ldots$. If $\sqrt{6} = 2 + 1/x$ then $x = \dfrac{1}{-2+\sqrt{6}} = \dfrac{-2-\sqrt{6}}{4 - 6} = \ldots$