Can someone help me find the continued fraction expansion for $\sqrt{7}$ just like I did for below.

For $\sqrt{3}$ I did this:

I was given that $x = \sqrt{3} -1 $

$x = \frac{1}{1+\frac{1}{2+x}} $

take the second

$x = \frac{1}{1+\frac{1}{2+x}} $

$x = \frac{1}{\frac{2+x+1}{2+x}} $

$x = \frac{2+x}{x+3} $

$x(x+3) = 2+x $

$x^2 +3x = 2+x \Rightarrow x^2+2x-2=0 $

find x,then we are done

how to write $x =\sqrt{3}-1$ in continued fraction using $x = \frac{1}{1+ \frac{1}{2+x}} $

$\sqrt{3}-1 = \dfrac{1}{1+ \dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\d frac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfr ac{1}{2+..}}}}}}}}}}$

I just sub x value each time it will never end

$\sqrt{3}-1 = \dfrac{1}{1+ \dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\d frac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfr ac{1}{2+..}}}}}}}}}} $

so

$\sqrt{3} = 1 +\dfrac{1}{1+ \dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\d frac{1}{2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfr ac{1}{2+..}}}}}}}}}} $

So it would be <1;1,2,1,2,1,2>