2. For the quadratic function $y=x^{2}+b x+c$, the vertex of its graph is $D$, and it intersects the positive x-axis at points $A$ and $B$ from left to right, and the positive y-axis at point $C$. If $\triangle A B D$ and $\triangle O B C$ are both isosceles right triangles (where $O$ is the origin), then $b+2 c=$
2. 2 .

From the given information, we have
$$
\begin{array}{l}
C(0, c) 、 A\left(\frac{-\dot{b}-\sqrt{b^{2}-4 c}}{2}, 0\right), \\
B\left(\frac{-b+\sqrt{b^{2}-4 c}}{2}, 0\right), D\left(-\frac{b}{2},-\frac{b^{2}-4 c}{4}\right) .
\end{array}
$$

Draw $D E \perp A B$ at point $E$. Then $2 D E=A B$,
which means $2 \times \frac{b^{2}-4 c}{4}=\sqrt{b^{2}-4 c}$.
Solving this, we get $\sqrt{b^{2}-4 c}=0$ (discard) or 2 .
$$
\begin{array}{l}
\text { Also, } O C=O B \\
\Rightarrow c=\frac{-b+\sqrt{b^{2}-4 c}}{2} \\
\Rightarrow b+2 c=\sqrt{b^{2}-4 c}=2 .
\end{array}
$$