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?

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.
