When the expression changes, the minimum value of the fraction $\frac{3 x^{2}+6 x-5}{\frac{1}{2} x^{2}+x+1}$ is $\qquad$. (1993, National Junior High School Competition)
Let $y=\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}$.
(*)
Rearranging (*), we get
$$
(y-6) x^{2}+(2 y-12) x+2 y-10=0 \text {. }
$$

Since $x$ is a real number, we have
$$
\Delta=(2 y-12)^{2}-4(y-6)(2 y-10) \geqslant 0 \text {, }
$$

which simplifies to $y^{2}-10 y+24 \leqslant 0$.
Solving this, we get $4 \leqslant y \leqslant 6$.
Substituting $y=4$ into (*) gives $x=-1$.
Thus, when $x=-1$, $y_{\text {min }}=4$.
Note: Generally, if a rational function $y=f(x)$ can be transformed into $p(y) x^{2}+q(y) x+r(y)=0$ after clearing the denominator, where $p, q, r$ are real functions of $y$, then the range of $y$ can be determined by $\Delta \geqslant 0$, and thus the extremum of $y$ can be found.
There is also a simpler solution for this problem:
$$
\begin{array}{l}
\frac{3 x^{2}+6 x+5}{\frac{1}{2} x^{2}+x+1}=\frac{\left(6 x^{2}+12 x+12\right)-2}{x^{2}+2 x+2} \\
=6-\frac{2}{(x+1)^{2}+1} .
\end{array}
$$

Clearly, when $x=-1$, the original expression has a minimum value of 4.