If $$\frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}= A+\sqrt{B}$$, then $$B-A$$ is

$$\sqrt{7 + 4\sqrt{3}} = \sqrt{7 + 2*2*\sqrt{3}}$$

= $$\sqrt{4 + 3 + 2*2*\sqrt{3}} = \sqrt{(2 + \sqrt{3})^2}$$

= $$2 + \sqrt{3}$$

Expression : $$\frac{4+3\sqrt{3}}{2 + \sqrt{3}}= A+\sqrt{B}$$

=> $$\frac{4 + 3\sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = A + \sqrt{B}$$

=> $$\frac{8 - 4\sqrt{3} + 6\sqrt{3} - 9}{4 - 3} = A + \sqrt{B}$$

=> $$2\sqrt{3} - 1 = A + \sqrt{B}$$

=> $$A = -1$$ and $$\sqrt{B} = 2\sqrt{3}$$

=> $$B = (2\sqrt{3})^2 = 12$$

=> $$B - A = 12 - (-1) = 13$$