1. The total number of integer solutions to the equation $5 x^{2}-6 x y+7 y^{2}=130$ is ( ) groups.
(A) 0
(B) 1
(C) 2
(D) 3
-.1.C.
Given the equation can be rearranged as
$$
5 x^{2}-6 x y+\left(7 y^{2}-130\right)=0 \text {. }
$$

From $\Delta>0$, we get $y^{2} \leqslant 25$.
Since $y$ is an integer, therefore, $-5 \leqslant y \leqslant 5$.
Upon inspection, $(3,5),(-3,-5)$ satisfy the conditions.