7. If $a$ is the positive root of the equation $x^{2}+3 x-2=0$, and $b$ is the root of the equation $x+\sqrt{x+1}=3$, then $a+b=$ $\qquad$

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7. 2 .

Notice that, $b+1+\sqrt{b+1}=4$.
Let $t=\sqrt{b+1}-1$.
Then $(t+1)^{2}+(t+1)=4$, and $t>0$.
Thus, $a$ satisfies $(a+1)^{2}+(a+1)=4$.
Therefore, $a=t$.
Hence $b+1+a+1=4 \Rightarrow a+b=2$.