Метод хорд

import sys
import math
def f(x):
    return 5*x**3-6*x
def f1(x):
    return 15*x**2-6
try:
    a=input(int('vvedite'))
    b=input(int('vvedite'))
    if a and b:
        i=int(a)
        j=int(b)
        while True:
                i=i-((f(a)*(j-i))/(f(j)-f(i)))
                j=j-f(j)/f1(j)
                print(j)

try:
    ...
except:
    ...
finally:
    ...
else:
    ...

def f(x:float) -> float:
    return 5*x**3-6*x
#--------------------------------
def secant(f, x0: float, eps: float=1e-7 ) -> float:
   """
   solves f(x) = 0 by secant method with precision eps
   :param f: f
   :param x0: starting point
   :param eps: precision wanted
   :return: root of f(x) = 0
   """
   x=x0
   x_prev=x0 + 2 * eps
   i = 0
   while abs(x - x_prev) >= eps and i < 1e3:
      x -= f(x) / (f(x) - f(x_prev)) * (x - x_prev)
      x_prev = x
      i += 1
   return x
#--------------------------------
x0=int(input("x="))
x=secant(f, x0)
print(x)