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?

$$2x^2-1=2[x]~~~~(1).$$ $[x]$ 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 $x=\sqrt{\frac{3}{2}}, \frac{1}{\sqrt{2}}$
