f(x) = ax^2 + bx + c
f(x)>-2x的解集為（1,3) => f(x) = -2x的解為1,3,且a c = 3a
1+3 = -(b+2)/a => b = -4a - 2
=> f(x) = ax^2 + bx + c = ax^2 - (4a+2)x + 3a
(1) 方程f（x)+6a=0有兩個(gè)相等的根
ax^2 - (4a+2)x + 3a + 6a = 0有兩個(gè)相等的根
=> a = 1
(2) f(x) = ax^2 - (4a+2)x + 3a = a(x - (4a+2)/2a)^2 +3a -(4a+2)^2/4a
f(x)的最大值為3a -(4a+2)^2/4a > 0
3a^2 - (2a+1)^2 < 0
-a^2 - 4a -1 < 0
a^2 + 4a + 1 > 0
a > -2 + 根號(hào)3 或 a < -2-根號(hào)3