Example 4 Find the range of real numbers $m$ such that the equation $x^{2}+(m-2) x+m+6=0$ about $x$:
(1) has one real root greater than 2 and the other real root less than 2;
(2) has two real roots both greater than 1.
Let $f(x)=x^{2}+(m-2) x+m+6$.
(1) According to the conditions, we have $f(2)<0 \\ f(1)>0 .\end{array}\right.$, so

that is
$$
\begin{array}{l}
\left\{\begin{array}{l}
(m-2)^{2}-4(m+6) \geqslant 0, \\
-\frac{m-2}{2}>1, \\
1^{2}+m-2+m+6>0,
\end{array}\right. \\
\left\{\begin{array}{l}
m \leqslant-2 \text { or } m \geqslant 10, \\
m-\frac{5}{2},
\end{array}\right.
\end{array}
$$

The range of real number $m$ is $-\frac{5}{2}<m \leqslant-2$.
Key Points for Solution This problem uses the method of constructing a quadratic function to solve the distribution of real roots of a quadratic equation. In (1), from $f(2)<0$ we get $\Delta \geqslant 0$.