# Solution.

If we represent an arbitrary number $x$ as $x=m+\alpha$, where $m$ is an integer, $\alpha \in(0,1)$, then $[x]=m=x-\alpha$. Thus, the original equation can be rewritten as

$$
x^{2}-x+\alpha=2019 \text {. }
$$

Since $\alpha \in(0,1)$, we have

$$
2018 \leq x^{2}-x \leq 2019
$$

Consider the function $f(x)=x^{2}-x$. This is a parabola with roots at 0 and 1 and branches directed upwards.

Olympiad for School Students "Hope of Energy". Final Stage. Solutions

Therefore, the obtained double inequality can be satisfied on some interval located to the left of the point $x=0$ and on another interval located to the right of the point $x=1$. Let's consider them in turn.

Suppose $x<0$. Then we need to find the interval where $f(x) \in [2018, 2019]$. We can estimate the values of $f(x)$ at integer points:

$$
\begin{aligned}
& f(-45)=(-45)(-44)=45^{2}-45=2025-45=1980 \\
& f(-44)=(-44)(-45)=45^{2}-45=2025-45=1980 \quad \text{(similarly calculated, also mentally)}
\end{aligned}
$$

Thus, $f(-45)=1980 < 2018$ and $f(-44)=1980 < 2018$. Therefore, the interval we are looking for is $x \in (-45, -44)$. In this case, $[x]=-45$ and the original equation becomes

$$
x^{2}+45=2019 \text {. }
$$

The solution (negative) is $x_{1}=-\sqrt{1974}$.

Now consider $x>0$. We need to find the interval where $f(x) \in [2018, 2019]$. We can estimate the values of $f(x)$ at integer points:

$$
\begin{aligned}
& f(44)=(44)(45)=45^{2}-45=2025-45=1980 \\
& f(45)=(45)^{2}-45=2025-45=1980 \quad \text{(similarly calculated, also mentally)}
\end{aligned}
$$

Thus, $f(44)=1980 < 2018$ and $f(45)=2025-45=1980 < 2018$. Therefore, the interval we are looking for is $x \in (45, 46)$. In this case, $[x]=45$ and the original equation becomes

$$
x^{2}-45=2019 \text {. }
$$

The solution (positive) is $x_{2}=\sqrt{2064}$.

Answer: $x_{1}=-\sqrt{1974}, x_{2}=\sqrt{2064}$.
# Problem 2.

Solve the equation

$$
x^{2}-[x]=2019,
$$

where $[x]$ denotes the integer part of $x$.

#