Example 6 Let $a, b, c$ be distinct real numbers, and satisfy the relation
$$
b^{2}+c^{2}=2 a^{2}+16 a+14,
$$

and $bc=a^{2}-4a-5$.
Find the range of values for $a$.
(2006, Hunan Province Junior High School Mathematics Competition)
From the given equations, we have
$$
\begin{array}{l}
(b+c)^{2} \\
=2 a^{2}+16 a+14+2\left(a^{2}-4 a-5\right) \\
=4(a+1)^{2} .
\end{array}
$$

Thus, \( b+c = \pm 2(a+1) \).
Therefore, we can construct a quadratic equation with roots \( b \) and \( c \):
$$
x^{2} \pm 2(a+1) x + a^{2} - 4 a - 5 = 0 .
$$

Since equation (1) has two distinct real roots, we have
$$
\Delta = 4(a+1)^{2} - 4\left(a^{2} - 4 a - 5\right) > 0 \text{. }
$$

Thus, \( a > -1 \).
When \( a > -1 \), we have
$$
\begin{array}{l}
b^{2} + c^{2} = 2 a^{2} + 16 a + 14 \\
= 2(a+1)(a+7) > 0 .
\end{array}
$$

Additionally, when \( a = b \), from equation (1) we have
$$
a^{2} \pm 2(a+1) a + a^{2} - 4 a - 5 = 0 \text{, }
$$

which simplifies to \( 4 a^{2} - 2 a - 5 = 0 \) or \( -6 a - 5 = 0 \).
Solving these, we get \( a = \frac{1 \pm \sqrt{21}}{4} \) or \( a = -\frac{5}{6} \).
Therefore, the range of \( a \) is
$$
a > -1 \text{ and } a \neq \frac{1 \pm \sqrt{21}}{4}, a \neq -\frac{5}{6} .
$$