We notice that
$$\sqrt{2010 + 2\sqrt{2009}} = \sqrt{1 + 2\sqrt{2009} + 2009} = \sqrt{(1 + \sqrt{2009})^2} = 1 + \sqrt{2009}.$$
Similarly, $\sqrt{2010 - 2\sqrt{2009}} = \sqrt{2009} - 1$. Since $1 + \sqrt{2009}$ is the solution of the quadratic equation $x^2 + a x + b = 0$, we have $2010 + 2\sqrt{2009} + a(1 + \sqrt{2009}) + b = 0$ or
$$\sqrt{2009}(2 + a) = -b - 2010 - a.$$
The right-hand side is an integer, so the left-hand side must be an integer as well. This implies $2 + a = 0$, thus $a = -2$. From $-b - 2010 - a = 0$ we get $b = -2008$. The two solutions of the quadratic equation $x^2 - 2x - 2008 = 0$ are $1 + \sqrt{2009}$ and $1 - \sqrt{2009}$. So $\sqrt{2009} - 1$ cannot be a solution.