二次函數(二)

#求 f(x)=y=3x**2-12x+10 y=0 x值多少 
import matplotlib.pyplot as plt
import numpy as np
import math
from sympy import Symbol,solve
x1=np.linspace(-6,10,100)
y1=3*x1**2-12*x1+10
plt.plot(x1,y1)
"""
 使用sympy無法求到解
x1=Symbol("x1")
f=Symbol("f")
f=3*x1**2-12*x1+10
ans=solve(f)
print(ans)
"""
#f(x)=y=3x**2-12x+10 a=3 b=-12 c=10
#x=(-b/2a)+math.sqrt(b**2-4ac)/2a
a=3
b=-12
c=10
x1=(-b/(2*a))+(math.sqrt(b**2-4*a*c)/(2*a))
#x1=(-b + (b**2-4*a*c)**0.5)/(2*a)
x2=(-b/(2*a))-(math.sqrt(b**2-4*a*c)/(2*a))
print(x1)
print(x2)
plt.plot(x1,0,"-o")
plt.text(x1,20,"("+str(round(x1,2))+","+str(0)+")")
plt.plot(x2,0,"-o")
plt.text(x2-2,20,"("+str(round(x2,2))+","+str(0)+")")
plt.show()