$解：??-\frac {1}{2}x2-4x- 6= 0??的解為??x_{1}= -6，???? x_{2}= -2??$
$??x2-6x+9=0??的解為??x_{1} =x_{2}= 3??$
$??x2-2x+3=0??無解$
$??y=-\frac {1}{2}x2- 4x- 6??與??x??軸的公共點(diǎn)為??(-6，???? 0)、???? (-2，???? 0)??$
$??y=x2-6x+9??與??x??軸的公共點(diǎn)為??(3 ，???? 0)??$
$??y=x2-2x+3??與??x??軸無公共點(diǎn)$
$二次函數(shù)與??x??軸有公共點(diǎn)，則公共點(diǎn)的橫坐標(biāo)就是方程的根;$
$二次函數(shù)與??x??軸無公共點(diǎn)，則方程無解。$
