解：对于方程$2x^2-4\sqrt {3}x + 3 = 0，$
这里$a = 2，$$b=-4\sqrt {3}，$$c = 3。$
$ $根据求根公式$x=\frac {-b\pm \sqrt {b^2-4ac}}{2a}，$
先计算判别式$∆=b^2-4ac$
$=(-4\sqrt {3})^2-4×2×3=48 - 24 = 24。$
$ $则$x=\frac {4\sqrt {3}\pm \sqrt {24}}{2×2}=\frac {4\sqrt {3}\pm 2\sqrt {6}}{4}=\frac {2\sqrt {3}\pm \sqrt {6}}{2}，$
所以$x_{1}=\frac {2\sqrt {3}+\sqrt {6}}{2}，$$x_{2}=\frac {2\sqrt {3}-\sqrt {6}}{2}。$
