smc

Dividing the quadratic by $7+4\sqrt{3}$ to obtain a monic polynomial will give a linear coefficient of $\frac{2+\sqrt{3}}{7+4\sqrt{3}}$. Rationalizing the denominator gives: $\frac{(2+\sqrt{3})(7-4\sqrt{3})}{7^2-4^2\cdot 3}$ $=\frac{14-12-\sqrt{3}}{49-48}$ $=2-\sqrt{3}$ Dividing the constant term by $7+4\sqrt{3}$ (and using the same radical conjugate as above) gives: $\frac{-2}{7+4\sqrt{3}}$ $=-2(7-4\sqrt{3})$ $=8\sqrt{3}-14$ So, dividing the original quadratic by the coefficient of $x^2$ gives $x^2+(2-\sqrt{3})x+8\sqrt{3}-14=0$ From the quadratic formula, the positive difference of the roots is $\frac{\sqrt{b^2-4ac}}{a}$. Plugging in gives: $\sqrt{(2-\sqrt{3}{)}^2-4(8\sqrt{3}-14)(1)}$ $=\sqrt{7-4\sqrt{3}-32\sqrt{3}+56}$ $=\sqrt{63-36\sqrt{3}}$ $=3\sqrt{7-4\sqrt{3}}$ Note that if we take $\frac{1}{3}$ of one of the answer choices and square it, we should get $7-4\sqrt{3}$. The only answers that are (sort of) divisible by $3$ are $6\pm 3\sqrt{3}$, so those would make a good first guess. And given that there is a negative sign underneath the radical, $6-3\sqrt{3}$ is the most logical place to start. Since $\frac{1}{3}$ of the answer is $2-\sqrt{3}$, and $(2-\sqrt{3}{)}^2=7-4\sqrt{3}$, the answer is indeed $\boxed{6-3\sqrt{3}}$.