5. The equation $\left(x^{2}-4\right)^{2}=x+4$ has ( ) real solutions.
(A) 1
(B) 2
(C) 3
(D) 4
5. D.

Let $x^{2}-4=y$.
Then $x+4=y^{2}$.
$$
\begin{array}{l}
\text { (1) }+ \text { (2) gives } \\
x^{2}+x=y^{2}+y \\
\Rightarrow(x-y)(x+y+1)=0 \\
\Rightarrow y=x \text { or } y=-x-1 .
\end{array}
$$

When $y=x$, substituting into equation (1) yields $x=\frac{1 \pm \sqrt{17}}{2}$;
When $y=-x-1$, substituting into equation (1) yields
$$
x=\frac{-1 \pm \sqrt{13}}{2} \text {. }
$$

Thus, the original equation has 4 real solutions.