Solving $x^2-3-\operatorname{frac}(x)=0$ 
Problem. Solve the equation
$$ x^2 - 3 - \operatorname{frac}(x) = 0 $$

I tried to solve this question by moving $3$ to the other side:
$$x^2 - \operatorname{frac}(x) = 3$$
Since the fractional part of R.H.S. is $0$,
$$
\operatorname{frac}\left(x^2\right) = \operatorname{frac}(x)
\qquad\text{or}\qquad
\operatorname{frac}\left(x^2\right) = 1 - \operatorname{frac}(x).$$
Since there is only 1 case for $\operatorname{frac}(x^2)=\operatorname{frac}(x)$ which is $\operatorname{frac}\left(x^2\right)=\operatorname{frac}(x)=0$ and no value of $x$ satisfies the equation, we can rule this condition out.
We are left with $\operatorname{frac}\left(x^2\right)=1-\operatorname{frac}(x)$ and $x$ is negative, but I am stuck here.
Can anyone please give me an idea to proceed with this question?
 A: Since $x^2 = 3 + \operatorname{frac}(x)$, we have $3 \leq x^2 < 4$. Solving this inequality, we have either
$$ \sqrt{3} \leq x < 2 \qquad\text{or} \qquad -2 < x \leq -\sqrt{3}. $$

*

*In the former case, we have $\lfloor x \rfloor = 1$ and hence we must have
$$ x^2 = 3 + (x - 1) = x + 2. $$
However, solving this equation gives $x = -1$ or $x = 2$, none of which satisfying the restriction $\sqrt{3} \leq x < 2$. So, there is no solution of the equation in this case.


*In the latter case, we have $\lfloor x \rfloor = -2$, and so, we have
$$ x^2 = 3 + (x + 2) = x + 5. $$
Solving this equation gives two values $x = \frac{1}{2}(1 \pm \sqrt{21})$, and indeed, the choice
$$ x = \bbox[color:navy;padding:5px;border:1px navy dotted;]{\frac{1}{2}(1-\sqrt{21})} \approx -1.79129 $$
satisfies the equation.
A: Here's a start:  Let $x = a + y$ where $a$ is an integer and $0\leq y <1$.  Try a few values for $a$, such as $0, 1, 2, 3$ and then $-1, -2, -3$.  I think you'll see how to proceed after that.
A: We have
$$
0 = x^2  - 3 - {\rm frac}(x) = x^2  - 3 - \left\{ x \right\}
$$
where
$$
x = \left\lfloor x \right\rfloor  + \left\{ x \right\}\quad \left| {\,0 \le \left\{ x \right\} < 1} \right.
$$
So we can put first of all
$$
\eqalign{
  & 3 \le x^2  = 3 + \left\{ x \right\} < 4\quad  \Rightarrow \quad \sqrt 3  \le  \pm x < 2\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {\sqrt 3  \le x < 2} \right) \cup \left( { - 2 < x \le  - \sqrt 3 } \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {1 \le \left\lfloor x \right\rfloor  < 2} \right)
 \cup \left( { - 3 < \left\lfloor x \right\rfloor  \le  - 2} \right)\quad  \Rightarrow \quad \left\lfloor x \right\rfloor
  =  - 2,1 \cr} 
$$
Then, for $\left\lfloor x \right\rfloor  = 1$
$$
\eqalign{
  & 0 = x^2  - 3 - \left\{ x \right\} = \left( {1 + \left\{ x \right\}} \right)^2  - \left\{ x \right\} - 3 =   \cr 
  &  = 1 + 2\left\{ x \right\} + \left\{ x \right\}^2  - \left\{ x \right\} - 3 =   \cr 
  &  = \left\{ x \right\}^2  + \left\{ x \right\} - 2\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {\left\{ x \right\} =  - 2} \right) \cup \left( {\left\{ x \right\} = 1} \right)
\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad {\rm no}\;{\rm solution} \cr} 
$$
while for $\left\lfloor x \right\rfloor  = -2$
$$
\eqalign{
  & 0 = x^2  - 3 - \left\{ x \right\} = \left( { - 2 + \left\{ x \right\}} \right)^2  - \left\{ x \right\} - 3 =   \cr 
  &  = 4 - 4\left\{ x \right\} + \left\{ x \right\}^2  - \left\{ x \right\} - 3 =   \cr 
  &  = \left\{ x \right\}^2  - 5\left\{ x \right\} + 1\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {\left\{ x \right\} = {{5 + \sqrt {21} } \over 2}} \right)
\cup \left( {\left\{ x \right\} = {{5 - \sqrt {21} } \over 2}} \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ x \right\} = {{5 - \sqrt {21} } \over 2} \cr} 
$$
That means that for $x$ we get
$$
x = \left\lfloor x \right\rfloor  + \left\{ x \right\}
 =  - 2 + {{5 - \sqrt {21} } \over 2} = {{1 - \sqrt {21} } \over 2}
$$
which is the same as the solution indicated by S. Lee
This is the graphical representation of the above

