2. Given that $k$ is an integer. If the quadratic equation $k x^{2}+(2 k+3) x+1$ $=0$ has rational roots, then the value of $k$ is $\qquad$
2. -2 .

From the given, $\Delta_{1}=(2 k+3)^{2}-4 k$ is a perfect square. Let $(2 k+3)^{2}-4 k=m^{2}(m$ be a positive integer), that is,
$$
4 k^{2}+8 k+9-m^{2}=0 \text {. }
$$

Considering equation (1) as a quadratic equation in $k$, by the problem's condition, it has integer roots, so the discriminant of equation (1)
$$
\Delta_{2}=64-16\left(9-m^{2}\right)=16\left(m^{2}-5\right)
$$

should be a perfect square.
Let $m^{2}-5=n^{2}(n$ be a positive integer, and $m>n)$, then we have $(m+n)(m-n)=5$.
Thus, $\left\{\begin{array}{l}m+n=5, \\ m-n=1 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}m=3, \\ n=2 .\end{array}\right.$
Substituting $m=3$ into equation (1) yields $k=-2$ or $k=0$ (discard).
Therefore, $k=-2$.