Continued fraction for $\sqrt{14}$ I'm referencing this page: An Introduction to the Continued Fraction, where they explain the algebraic method of solving the square root of $14$.
$$\sqrt{14} = 3 + \frac1x$$
So, $x_0 = 3$, Solving for $x$, we get
$$x = \frac{\sqrt{14} + 3}5$$
However, in the next step, how do we get the whole number $x_1$ = 1?
$$\frac{\sqrt{14} + 3}5 = 1 + \frac1x$$
My understanding is we would substitute for $x$ in the original equation for $\sqrt{14}$ whereas $$\sqrt{14} = 3 + \frac1{\frac{\sqrt{14} + 3}5}$$
Then substitute the $\sqrt{14}$ again here for $x = \frac{\sqrt{14} + 3}5$ to get the $x_1$ of the continued fraction? Am I just getting the algebra at this point wrong or am I botching steps? 

$\sqrt{14}=3+\sqrt{14}-3=3+\frac{1}{\frac{\sqrt{14}+3}{5}}\implies x_0 = 3$
$\frac{\sqrt{14}+3}{5}=\frac{6+\sqrt{14}-3}{5}=1+\frac{\sqrt{14}-2}{5}=1+\frac{1}{\frac{\sqrt{14}+2}{2}} \implies x_1 = 1$
$\frac{\sqrt{14}+2}{2}=\frac{5+\sqrt{14}-3}{2}=2+\frac{\sqrt{14}-2}{2}=2+\frac{1}{\frac{\sqrt{14}+2}{5}} \implies x_2 = 2$
etc.
