Example 10 Let real numbers $x, y, z$ satisfy
$$
x+y+z=5, \quad xy+yz+zx=3 \text{.}
$$

Then the maximum value of $z$ is $\qquad$.
Solve: From $x+y+z=5$, we get $x+y=5-z$.
Let $x=\frac{5-z}{2}+t, y=\frac{5-z}{2}-t$.
Substituting into $x y+y z+z x=3$ and simplifying, we get $3 z^{2}-10 z-13=-4 t^{2}$.
Therefore, $3 z^{2}-10 z-13 \leqslant 0$.
Solving, we get $-1 \leqslant z \leqslant \frac{13}{3}$.
Thus, the maximum value of $z$ is $\frac{13}{3}$.