I've been thinking about repeating square roots: $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$. I wrote a program on my calculator to do it $n$ times and I found that, if $x = y^2 - y$ then $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ approaches $y$ as $n$ gets large. So, solving for $y$:
$$y^2 - y - x = 0$$ $$y = \frac{1 \pm \sqrt{1 + 4x}}{2} = \frac{\sqrt{4x + 1} + 1}{2}$$
$$\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} =\frac{\sqrt{4x + 1} + 1}{2} $$
This works, but where does the $y^2-y=x$ come from? I got here via plugging and guessing, but how can I prove that with an input of $x$ where $x = y^2 - y$, the answer will be $y$?
If we let $$y = \sqrt{x+\color{red}{\sqrt{x+\sqrt{x + \cdots}}}},$$ then if the limit were to exist, the inner square-root in $\color{red}{\text{red}}$ is also $y$ and hence we have $$y = \sqrt{x+y}$$ Squaring the above, we get that $$y^2 = x+y$$from which we can find $y$ as you have done. The slightly non-trivial part though is proving that $$\sqrt{x+\color{black}{\sqrt{x+\sqrt{x + \cdots}}}}$$ makes sense, i.e., the limit exists.
EDIT
To prove the above limit exists, here is one way.
$1$. Define $y_1 = \sqrt{x}$ and $y_{n+1} = \sqrt{x+y_n}$.
$2$. Prove by induction, that $$ \sqrt{x} < y_n<\dfrac{1+\sqrt{4x+1}}2$$
$3$. Make use of $(2)$ and prove by induction that $y_n$ is an increasing sequence.
$4$. Recall from real analysis that $\mathbb{R}$ is complete (i.e., every bounded monotone sequence converges to prove that a limit exists)