ax2 + bx + c = 0 的两个解。

((-b + math.sqrt(s))/(2*a), (-b - math.sqrt(s))/(2*a))

# 解的个数
s = b*b - 4*a*c
# 0解
if s < 0:
print("无解")
return
# 1解
elif s == 0:
print("有唯一解: %.2f" % -b/(2*a))
# 2解
else:
print("有两个解: %.2f, %2.f" %((-b + math.sqrt(s))/(2*a), (-b - math.sqrt(s))/(2*a)))