232. $\frac{1}{12} \lg ^{2} x=\frac{1}{3}-\frac{1}{4} \lg x$.
Solution. This is a logarithmic equation reducible to a quadratic. By setting $\lg x=z$, we obtain the equation

$$
\frac{1}{12} z^{2}=\frac{1}{3}-\frac{1}{4} z \text { or } z^{2}+3 z-4=0
$$

Here $a=1, b=3, c=-4, \quad D=b^{2}-4 a c=9-4 \cdot 1(-4)=9+16=25$; $\sqrt{D}=5$. Using the formula $z=\frac{-b \pm \sqrt{D}}{2 a}$, we find $z_{1}=\frac{-3-5}{2}=$ $=-4, z_{2}=\frac{-3+5}{2}=1$.

Since $\lg x=z$, then $\lg x=-4$ or $\lg x=1$. Therefore, $x_{1}=$ $=10^{-4}=0.0001, x_{2}=10$.