Solution:

The answer is 414. From the usual formula for solving quadratic equations, we get $x=2-\sqrt{2}$. Now using the formula for the sum of a geometric progression, we have
$$\begin{aligned} x+x^{2}+x^{3}+\cdots+x^{2009} & =x\left(1+x+x^{2}+\cdots+x^{2008}\right) \\ & =x \frac{1-x^{2009}}{1-x}=\frac{x}{1-x}-\frac{x^{2010}}{1-x} . \end{aligned}$$
The second term of this sum is very small: indeed, with very loose estimates, we obtain
$$(2-\sqrt{2})^{2010}<(0.6)^{2010}<\left((0.6)^{2}\right)^{1005}<\frac{1}{2^{1005}}<\frac{1}{\left(2^{10}\right)^{100}}<\frac{1}{10^{100}} .$$
On the other hand, the first term equals
$$\frac{x}{1-x}=\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{(2-\sqrt{2})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{\sqrt{2}}{1} .$$
Therefore the value of the expression is a quantity that differs from $\sqrt{2}$ by less than $10^{-100}$; the first digits of its decimal expansion will therefore be the same as those of $\sqrt{2}$, that is $1.414 \ldots$.