Let $x$- minimal root of equation $x^2-4x+2=0$. 
Find two first digits of number $ \{x+x^2+....+x^{20} \}$ after $0$, where $\{a\}$- fractional part of $a$.
1. **Find the minimal root of the equation \(x^2 - 4x + 2 = 0\):**
   The roots of the quadratic equation \(ax^2 + bx + c = 0\) are given by the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   For the equation \(x^2 - 4x + 2 = 0\), we have \(a = 1\), \(b = -4\), and \(c = 2\). Plugging these values into the quadratic formula, we get:
   \[
   x = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}
   \]
   The minimal root is:
   \[
   x = 2 - \sqrt{2}
   \]

2. **Express the sum \(x + x^2 + \cdots + x^{20}\) in a closed form:**
   The sum of a geometric series \(S = x + x^2 + \cdots + x^{20}\) can be written as:
   \[
   S = \sum_{k=1}^{20} x^k = x \frac{x^{20} - 1}{x - 1}
   \]
   Substituting \(x = 2 - \sqrt{2}\), we get:
   \[
   S = (2 - \sqrt{2}) \frac{(2 - \sqrt{2})^{20} - 1}{(2 - \sqrt{2}) - 1} = (2 - \sqrt{2}) \frac{(2 - \sqrt{2})^{20} - 1}{1 - \sqrt{2}}
   \]

3. **Simplify the expression:**
   Let \(A = (2 - \sqrt{2})^{20}\). Then:
   \[
   S = (2 - \sqrt{2}) \frac{A - 1}{1 - \sqrt{2}}
   \]
   We need to find the fractional part of \(S\). First, we approximate \(A\):
   \[
   2 - \sqrt{2} \approx 0.5858
   \]
   Since \((2 - \sqrt{2})^{20}\) is a very small number, we can approximate:
   \[
   A \approx 0
   \]
   Therefore:
   \[
   S \approx (2 - \sqrt{2}) \frac{-1}{1 - \sqrt{2}} = (2 - \sqrt{2}) \frac{-1}{1 - \sqrt{2}} = (2 - \sqrt{2}) \frac{-1}{1 - \sqrt{2}} = \sqrt{2} - (2 - \sqrt{2})^{21}
   \]

4. **Estimate the bounds of \(S\):**
   Since \(2 - \sqrt{2} < \frac{2}{3}\), we have:
   \[
   (2 - \sqrt{2})^{20} < \left(\frac{2}{3}\right)^{20}
   \]
   \[
   \left(\frac{2}{3}\right)^{20} \approx 2^{-20} \approx 9^{-10} \approx 2^{-30}
   \]
   Therefore:
   \[
   \sqrt{2} > S > \sqrt{2} - 2^{-9}
   \]
   \[
   \sqrt{2} \approx 1.414
   \]
   \[
   1.414 > S > 1.414 - 0.002 = 1.412
   \]

5. **Determine the fractional part:**
   The fractional part of \(S\) is:
   \[
   \{S\} = S - \lfloor S \rfloor
   \]
   Since \(1.412 < S < 1.414\), we have:
   \[
   \{S\} = S - 1
   \]
   \[
   0.412 < \{S\} < 0.414
   \]

The final answer is \(\boxed{41}\).