jmc

Since the root $\sqrt{3}-2$ is irrational but the coefficients of the quadratic are rational, from the quadratic formula we can see that the other root must be $-\sqrt{3}-2.$

To find the quadratic, we can note that the sum of the roots is $\sqrt{3}-2-\sqrt{3}-2=-4$ and the product is $(\sqrt{3}-2)(-\sqrt{3}-2)=4-3=1.$ Then by Vieta's formulas, we know that the quadratic $\boxed{x^2+4x+1}$ has $\sqrt{3}-2$ as a root.