Simple square root calculator

The great input from @hjpotter92 gets you 99% of the way there. The only additional comment I have is about this:

while num != random_num * random_num:

Exact comparison of floating-point numbers is problematic. You're going to want to pick some (very small) number, usually called epsilon, below which the error is insignificant and call that "close enough".

Read https://docs.python.org/3/library/math.html#math.isclose for more details.

Welcome to the community :). A few pointers:

1. The code does not follow PEP-8 style guide. You should follow the snake_case naming convention for variables.
2. Split code into separate functions.
3. Instead of the code running as is, the execution condition should be placed inside an if __name__ block.
4. If you're on python 3.6+, make use of f-strings over concatenating separate variables and casting them to strings over and over.
5. Cast the input from user as float as soon as it is read.
6. I don't think you need to round the random variable calculated to 4 digits. For print purpose, you can use the precision specifier.
7. Again, if using python 3.6+, you can also provide type hinting.

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The above mentioned points, and a few more changes, your code would be:

from math import isclose
from random import randint


def calculate_sq_root(num: float) -> float:
    "generate a random number and manipulate it until we calculate square root of input number"
    random_num = randint(1, 9)
    print(f"Random number is {random_num}.")
    while not isclose(num, random_num * random_num):
        random_num = (random_num ** 2 + num) / (random_num * 2)
        print(f"Trying number {random_num:.4f}")
    return random_num


def get_input() -> float:
    "ask user for a number to find sq root for"
    return float(input("Please enter number for which we'll find sq root:n"))


def main():
    user_num = get_input()
    sq_root = calculate_sq_root(user_num)
    print(f"Square root of {user_num} is {sq_root}.")


if __name__ == "__main__":
    main()