What quadratic polynomial whose coeﬃcient of $x^2$ is $1$ has roots which are the complex conjugates of the solutions of $x^2 -6x+ 11 = 2xi-10i$? (Note that the complex conjugate of $a+bi$ is $a-bi$, where a and b are real numbers.)
1. **Identify the given quadratic equation and its roots:**
   The given quadratic equation is:
   \[
   x^2 - 6x + 11 = 2xi - 10i
   \]
   To find the roots, we first rewrite the equation in standard form:
   \[
   x^2 - 6x + 11 - 2xi + 10i = 0
   \]
   This can be written as:
   \[
   x^2 - (6 + 2i)x + (11 + 10i) = 0
   \]
   Let the roots of this equation be \(\alpha\) and \(\beta\).

2. **Find the polynomial with roots that are the complex conjugates of \(\alpha\) and \(\beta\):**
   The complex conjugates of \(\alpha\) and \(\beta\) are \(\overline{\alpha}\) and \(\overline{\beta}\). The polynomial with these roots is:
   \[
   (x - \overline{\alpha})(x - \overline{\beta})
   \]
   Using the properties of complex conjugates, we know:
   \[
   \overline{\alpha + \beta} = \overline{\alpha} + \overline{\beta}
   \]
   and
   \[
   \overline{\alpha \beta} = \overline{\alpha} \cdot \overline{\beta}
   \]

3. **Calculate the sum and product of the roots:**
   Given:
   \[
   \alpha + \beta = 6 + 2i
   \]
   and
   \[
   \alpha \beta = 11 + 10i
   \]
   Therefore:
   \[
   \overline{\alpha + \beta} = \overline{6 + 2i} = 6 - 2i
   \]
   and
   \[
   \overline{\alpha \beta} = \overline{11 + 10i} = 11 - 10i
   \]

4. **Form the quadratic polynomial with the conjugate roots:**
   The polynomial with roots \(\overline{\alpha}\) and \(\overline{\beta}\) is:
   \[
   x^2 - (\overline{\alpha + \beta})x + \overline{\alpha \beta}
   \]
   Substituting the values:
   \[
   x^2 - (6 - 2i)x + (11 - 10i)
   \]

The final answer is \(\boxed{x^2 - (6 - 2i)x + (11 - 10i)}\).