3. If the constant $a$ makes the equation about $x$
$$
\lg \left(x^{2}+20 x\right)-\lg (8 x-6 a-3)=0
$$

have a unique solution. Then the range of values for $a$ is $\qquad$
3. $-\frac{163}{6}0$, that is
$$
\left\{\begin{array}{l}
x(x+20)>0, \\
x^{2}+12 x+6 a+3=0 .
\end{array}\right.
$$

From (2), solve for $x_{1}=-6-\sqrt{33-6 a}, x_{2}=-6+\sqrt{33-6 a}$. At this point, there are two possible scenarios:
(1) $\left\{\begin{array}{l}-6-\sqrt{33-6 a}0 .\end{array}\right.$
(1) has no solution, while from (2) we get $-\frac{163}{6}<a<-\frac{1}{2}$.
Note: This problem can also be solved using the "graphical method".