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?
 A: Hint
$$2x^2-1=2\lfloor x \rfloor \leq 2x$$
Therefore 
$$2x^2-2x-1 \leq 0$$
Solve, and the solution will tell you that there are only few potential values for $\lfloor x \rfloor$. Solve for each.
A: 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 $n \ge 2$,  $\sqrt{n + \dfrac 12} - n < 0$
A: $$2x^2-1=2[x]~~~~(1).$$ $[x]$ is GIF/ Floor function
We have $2[x]+1=2x^2 \ge 0 \implies [x] \ge 0 ~~~(2)$
Let $x=n+q$, $n$-is integer $n\ge 0$ and $0 \le q <1$, then we get 
$$2n^2+2q^2+4nq-1=2n~~~~(3)$$
$$\implies 2n^2-2n-1=-2q^2-4nq \le 0 \implies 2n^2-2n-1 \le 0 \implies  \frac{1-\sqrt{3}}{2} \le  n \le \frac{1+\sqrt{3}}{2}.$$
$$\implies n=0,1$$
Now put $n=0$ in (3) to get $q=\frac{\sqrt{6}-2}{2} >0, \implies x =1+q =\sqrt{\frac{3}{2}}.$
Next put $n=0$ in (3) to get $2q^2-1=0 \implies q=\frac{1}{\sqrt{2}}>0 \implies x=0+q=\frac{1}{\sqrt{2}}.$
So only two solutions $x=\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}$
