Find the third proportional of $ (3 +\sqrt{2})$ and $ 2\sqrt{7} $

Let the third proportional to $ (3+\sqrt{2}) \text{ and } 2\sqrt{7} \text{ be } x.$
By definition: $ \dfrac{(3+\sqrt{2})}{2\sqrt{7}} = \dfrac{2\sqrt{7}}{x} $
$ x = \dfrac{(2\sqrt{7})(2\sqrt{7})}{(3+\sqrt{2})} $
$x = \dfrac{28}{(3+\sqrt{2})} $
$ x = \dfrac{28(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})} $
$ x = \dfrac{28(3-\sqrt{2})}{9-2} $
$ x = \dfrac{28(3-\sqrt{2})}{7} $
$ x = 4(3-\sqrt{2}) $
$ x = 12 - 4\sqrt{2}$
$12 - 4\sqrt{2}$
$ 4(3 - \sqrt{2})$