jmc

Because the coefficients of the polynomial are rational, the radical conjugate $2-\sqrt{3}$ must also be a root of the polynomial. By Vieta's formulas, the product of the roots of this polynomial is $-10,$ and the product of these two roots is $(2+\sqrt{3})(2-\sqrt{3})=1,$ so the remaining root must be $\frac{-10}{1}=-10.$ Then by Vieta's formulas again, we have $$b=(-10)(2-\sqrt{3})+(-10)(2+\sqrt{3})+(2+\sqrt{3})(2-\sqrt{3})=\boxed{-39}.$$