3. The range of the function $y=\lg \left(x^{2}+2 x-a\right)$ is $\mathbf{R},$ the range of values for $a$ is
3. $a \geqslant-1$.

Let $t=x^{2}+2 x-a$, then according to the problem, the range of $t$ should include $(0,+\infty), t=(x+1)^{2}-a-1$, hence $-a-1 \leqslant 0$, which gives $a \geqslant-1$.