Example 9 Given the equation
$$
x^{2}-11 x+30+m=0
$$

has two real roots both greater than 5. Find the range of values for $m$.


Let $y=x-5$, then the original equation is equivalently transformed into
$$
y^{2}-y+m=0 \text {. }
$$

Thus, the two real roots of the original equation being both greater than 5 is equivalent to equation (1) having two positive real roots.
Therefore, according to the theorem, we have
$$
\left\{\begin{array}{l}
\Delta=1-4 m \geqslant 0, \\
y_{1}+y_{2}=1>0, \\
y_{1} y_{2}=m>0 .
\end{array}\right.
$$

Solving the system of inequalities, we get $0<m \leqslant \frac{1}{4}$.