解：對于方程?$2x^2-4\sqrt {3}x + 3 = 0，$?
這里?$a = 2，$??$b=-4\sqrt {3}，$??$c = 3。$?
?$ $?根據(jù)求根公式?$x=\frac {-b\pm \sqrt {b^2-4ac}}{2a}，$?
先計(jì)算判別式?$?=b^2-4ac$?
?$=(-4\sqrt {3})^2-4×2×3=48 - 24 = 24。$?
?$ $?則?$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}。$?
