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$\ds{\bbox[5px,#ffd]{x^{2} - 20\left\lfloor\,{x}\,\right\rfloor +
19 = 0}\,,\quad x = {\Large ?}}$
It is clear that
$\ds{\left\lfloor x\right\rfloor = {x^{2} + 19 \over 20} = m,
\mbox{where}\ m \in \mathbb{N}_{\geq\ 1}\ \mbox{such that}\ x = \root{20m - 19}}$.
Then,
\begin{align}
&\bbox[5px,#ffd]{m = \left\lfloor\,{\root{20m - 19}}\,\right\rfloor}
\implies m \leq \root{20m - 19} < m + 1
\\[5mm] & \implies m^{2} \leq 20m - 19 < m^{2} + 2m + 1
\implies
\left\{\begin{array}{lcl}
\ds{m^{2} - 20m + 19} & \ds{\leq} & \ds{0}
\\
\ds{m^{2} - 18m + 20} & \ds{>} & \ds{0}
\end{array}\right.
\\[5mm] &\
\mbox{with solutions}\quad
1 \leq m <\ \underbrace{9 - \root{61}}_{\ds{\approx 1.1898}}\
\quad\mbox{or}\quad
\underbrace{9 + \root{61}}_{\ds{\approx 16.8102}}\ < m \leq 19
\\[5mm] &\
\implies m \in \braces{1,17,18,19} \implies
\bbx{x \in \braces{1,\root{321},\root{341},19}} \\ &
\end{align}
with $\ds{\root{321} \approx 17.9165}$ and
$\ds{\root{341} \approx 18.4662}$. Please, check for $\ds{\color{red}{x < 0}}$.
Best Answer
The solutions of the given equation are the solutions of the following system of inequalities $$\begin{cases} x^2-5 x<-5\\ x^2-5 x\geq -6\\ \end{cases} $$ That is $$\frac{1}{2} \left(5-\sqrt{5}\right)<x\leq 2\lor 3\leq x<\frac{1}{2} \left(\sqrt{5}+5\right)$$