How to solve $2x^2-2\lfloor x \rfloor-1=0$ How do I solve $2x^2-2\lfloor x \rfloor-1=0$?
I have tried setting $x=\lfloor x \rfloor + \{x\}$, where $\{x\}$ is the fractional part of $x$. Then, I tried $$2x^2-2\lfloor x \rfloor-1=0$$ $$2(\lfloor x \rfloor + \{x\})^2-2\lfloor x \rfloor-1=0$$ $$2\lfloor x \rfloor^2 + 4\lfloor x \rfloor\{x\}+2\{x\}^2-2\lfloor x \rfloor-1=0$$ but now I am stuck. How should I proceed?

Suppose $x = n + \delta$ where $n$ is an integer and $0 \le \delta < 1$ 
Then 
\begin{align}
   2x^2-2\lfloor x \rfloor-1&=0 \\
   x^2 &= \lfloor x \rfloor + \dfrac 12 \\
   x &= \sqrt{\lfloor x \rfloor + \dfrac 12} \\
   n+\delta &= \sqrt{n + \dfrac 12} \\
   \delta &= \sqrt{n + \dfrac 12} - n \\
\end{align}
\begin{array}{|c|c|c|}
\hline
   n & \delta & x \\
\hline
   0 & \dfrac{\sqrt 2}{2} & \dfrac{\sqrt 2}{2} \\
   1 & \dfrac{\sqrt 6}{2}-1 & \dfrac{\sqrt 6}{2} \\
   2 & \color{red}{\dfrac{\sqrt{10}}{2}-2 < 0} \\
 \hline
\end{array}
We have to stop at $n=2$ because, for $n \ge 2$,  $\sqrt{n + \dfrac 12} - n < 0$
