2. Given $m=1+\sqrt{2}, n=1-\sqrt{2}$, and
$$
\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8 \text {. }
$$

Then the value of $a$ is ( ).
(A) -5
(B) 5
(C) -9
(D) 9
2.C.

From the given, we have $m^{2}-2 m=1, n^{2}-2 n=1$.
Also, $\left(7 m^{2}-14 m+a\right)\left(3 n^{2}-6 n-7\right)=8$, so, $(7+a)(3-7)=8$.
Solving for $a$ gives $a=-9$.