1. Given the parabola
$$
y=x^{2}+(k+1) x+1
$$

intersects the $x$-axis at two points $A$ and $B$, not both on the left side of the origin. The vertex of the parabola is $C$. To make $\triangle A B C$ an equilateral triangle, the value of $k$ is $\qquad$
$=1 .-5$.
From the problem, we know that points $A$ and $B$ are to the right of the origin, and
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{(k+1)^{2}-4} \text {. } \\
\text { Then } \frac{\sqrt{3}}{2} \sqrt{(k+1)^{2}-4}=\left|1-\left(\frac{k+1}{2}\right)^{2}\right| \text {. }
\end{array}
$$

Solving this, we get $k=-5$.