1. [ $x$ ] represents the greatest integer not greater than $x$, then the real solution $x$ of the equation $\frac{1}{2} \times\left[x^{2}+x\right]=19 x+99$ is $\qquad$.
$$
=1 .-\frac{181}{38} \text { or } \frac{1587}{38} \text {. }
$$

From the condition, we get $38 x+198=\left[x^{2}+x\right] \in \mathbb{Z}$. Therefore, $38 x \in \mathbb{Z}$. Let $38 x=k \in \mathbb{Z}$, then $x=\frac{k}{38}(k \in \mathbb{Z})$. Substituting into the equation, we get $\left[\frac{k^{2}+38 k}{38^{2}}\right]=k+198$, which is equivalent to the system of inequalities $(k \in \mathbb{Z})$:
$$
\frac{k^{2}+38 k}{38^{2}}-1<k+198 \leqslant \frac{k^{2}+38 k}{38^{2}} \text {. }
$$

That is, $k^{2}+38 k-38^{2}<38^{2} k+198 \times 38^{2} \leqslant k^{2}+38 \times k$, or equivalently, $\left\{\begin{array}{l}k^{2}-38 \times 37 \times k-199 \times 38^{2}<0, \\ k^{2}-38 \times 37 \times k-198 \times 38^{2} \geqslant 0 .\end{array}\right.$

Thus, $19 \times(37-\sqrt{2165})<k<19 \times(37+\sqrt{2165})$ or $k \leqslant 19 \times(37-\sqrt{2161})$
or $k \geqslant 19 \times(37+\sqrt{2161})$.
$$
\begin{aligned}
\text { Therefore, } & -181.06 \cdots \approx 19 \times(37-\sqrt{2165})<k \\
& \leqslant 19 \times(37-\sqrt{2161}) \approx-180.24 \cdots
\end{aligned}
$$

or $1586.24 \cdots \approx 19 \times(37+\sqrt{2161}) \leqslant k$
$$
<19 \times(37+\sqrt{2165}) \approx 1587.06 \text {. }
$$

Therefore, $k=-181$ or $k=1587$.
Hence, $x=-\frac{181}{38}$ or $\frac{1587}{38}$.