（1）f(0)>0 => c>0;
f(1)>0 => 3a+2b+c>0;
2a+2b+2c=0;
so
a-c>0 => a>c>0;
c=-a-b => 2a+b>0 =>2a>-b>0 => b (b/a)>-2
c=-a-b>0 => a+b0) =>(b/a)0 => 有两个不等实根
-2