One. (20 points) Consider the equation in $x$:
$$
x^{2}-8 x+a|x-4|+12-2 a=0
$$

which has exactly two real roots. Find the range of values for $a$.
$$
\begin{array}{l}
|x-4|^{2}+a|x-4|-2(a+2)=0 . \\
\text { Let } y=|x-4|(y \geqslant 0) . \\
\text { Then } y^{2}+a y-2(a+2)=0 \\
\Rightarrow[y+(a+2)](y-2)=0 \\
\Rightarrow y_{1}=-(a+2), y_{2}=2,
\end{array}
$$

That is, $|x-4|=-(a+2)$ or $|x-4|=2$.
Therefore, the original equation has exactly two real roots
$$
\Leftrightarrow-(a+2)=2 \text { or }-(a+2)<0.
$$