Example 4 Given that $a, b, c$ satisfy $a+b+c=0, abc=8$. Then the range of values for $c$ is $\qquad$
Given the conditions, we have
$$
a+b=-c, a b=\frac{8}{c} \text {. }
$$

By the inverse of Vieta's formulas, $a, b$ are the two real roots of the equation
$$
t^{2}+c t+\frac{8}{c}=0
$$

Applying the discriminant theorem, we get
$$
\Delta=c^{2}-4 \times \frac{8}{c} \geqslant 0 .
$$

Solving for $c$, the range of $c$ is $c<0$ or $c \geqslant$
$$
2 \sqrt[3]{4}
$$