10.1. Form a quadratic equation with integer coefficients, one of whose roots is the number $2-\sqrt{7}$.
Answer. $x^{2}-4 x-3=0$.

Solution. If $a-\sqrt{b}$ is a root of the equation $x^{2}+p x+q$, then $a+\sqrt{b}$ is also a root of this equation. Indeed, by Vieta's theorem, the sum of the roots is equal to $-p$ (an integer), which means the second root has the form $c+\sqrt{b}$, and their product is also equal to the integer $q$, from which it follows that $c=a$. Thus, the equation has roots $2 \pm \sqrt{7}$. By Vieta's theorem, $p=-4, q=-3$.

Comment. A correct answer (including one found by trial and error), for which it is shown in any way that it satisfies the condition - 7 points. In this solution, the points are distributed as follows: it is justified that the equation has roots $2 \pm \sqrt{7}$ - 5 points, the coefficients of the equation are found - 2 points; the points are summed. Only the final equation is given without justification and verification - 0 points. The problem is not solved or solved incorrectly - 0 points.