7. The eccentricity of the conic section $6 x^{2}+4 x y+9 y^{2}=20$ is $\qquad$ .
7. $\frac{\sqrt{2}}{2}$.

The original equation is transformed into
$$
\begin{array}{c}
\left(4 x^{2}-4 x y+y^{2}\right)+\left(2 x^{2}+8 x y+8 y^{2}\right)=20 \\
\Rightarrow(2 x-y)^{2}+2(x+2 y)^{2}=20 \\
\Rightarrow \frac{\left(\frac{2 x-y}{\sqrt{5}}\right)^{2}}{4}+\frac{\left(\frac{x+2 y}{\sqrt{5}}\right)^{2}}{2}=1 .
\end{array}
$$

Notice that, $\frac{|2 x-y|}{\sqrt{5}}$ represents the distance from the point $(x, y)$ to the line $2 x-y=0$, and $\frac{|x+2 y|}{\sqrt{5}}$ represents the distance from the point $(x, y)$ to the line $x+2 y=0$. Since the two lines are perpendicular, if they are used as the axes of a coordinate system, the equation of the quadratic curve becomes $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$, which is an ellipse with $a=2, b=\sqrt{2}$.
Thus, the eccentricity $e=\frac{c}{a}=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\frac{\sqrt{2}}{2}$.