2. Let real numbers $x, y$ satisfy the equation $2 x^{2}+3 y^{2}=6 y$. Then the maximum value of $x+y$ is $\qquad$ .
2. $1+\frac{\sqrt{10}}{2}$.

Let $x+y=t$. Then $x=t-y$.
From $2(t-y)^{2}+3 y^{2}=6 y$, we get
$5 y^{2}-2(2 t+3) y+2 t^{2}=0$.
Since $y$ is a real number, therefore,
$$
\begin{array}{l}
\Delta=4(2 t+3)^{2}-4 \times 5 \times 2 t^{2} \geqslant 0 \\
\Rightarrow \frac{2-\sqrt{10}}{2} \leqslant t \leqslant \frac{2+\sqrt{10}}{2}
\end{array}
$$

Thus, the maximum value of $x+y$ is $1+\frac{\sqrt{10}}{2}$.