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quadratic equation formula

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YOU.com answered on June 2, 2021 Popularity 10/10 Helpfulness 2/10

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quadratic equation formula

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import math

def solve_quadratic_equation(a, b, c):
    discriminant = b**2 - 4*a*c
    
    if discriminant > 0:
        # Two distinct real roots
        root1 = (-b + math.sqrt(discriminant)) / (2*a)
        root2 = (-b - math.sqrt(discriminant)) / (2*a)
        return root1, root2
    elif discriminant == 0:
        # One real root
        root = -b / (2*a)
        return root
    else:
        # Complex roots
        real_part = -b / (2*a)
        imaginary_part = math.sqrt(abs(discriminant)) / (2*a)
        root1 = complex(real_part, imaginary_part)
        root2 = complex(real_part, -imaginary_part)
        return root1, root2

# Example usage with coefficients a, b, and c
a = 1
b = -3
c = 2

roots = solve_quadratic_equation(a, b, c)
print("Roots:", roots)

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Contributed on Sep 12 2023

YOU.com

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quadratic formula

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113

 (-b±√(b²-4ac))/(2a)

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Contributed on Jun 02 2021

Obedient Oryx

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quadratic formula

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x = [-b ± √(b² - 4ac)] / (2a)

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Contributed on Oct 18 2021

LoneWolf

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quadratic equation

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(-b-sqrt(b**2-4*a*c))/(2*a)
(-b+sqrt(b**2-4*a*c))/(2*a)
If that doesn't work,
((0-b)-sqrt(b**2-4*a*c))/(2*a)
((0-b)+sqrt(b**2-4*a*c))/(2*a)

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Contributed on Dec 31 2021

Thankful Toucan

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quadratic equations

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Formula:
(-b±√(b²-4ac))/(2a)

general form:
ax^2 + bx + c = 0
x^2 + bx + c = 0

some of the other hidden forms:
x^2 = bx - c (quadriatic after transposition)
c/bx+d = x (where you end up with a quadriatic after cross multiplication)

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Contributed on Feb 11 2023

Al Ayman Sarker

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quadratic equation calculator

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import tkinter as tk
import cmath

root= tk.Tk()

root.title("Quadratic Solver")

canvas1 = tk.Canvas(root, width = 400, height = 300)
canvas1.pack()

entry1 = tk.Entry (root) 
canvas1.create_window(200, 60, window=entry1)

entry2 = tk.Entry (root) 
canvas1.create_window(200, 100, window=entry2)

entry3 = tk.Entry (root) 
canvas1.create_window(200, 140, window=entry3)


def solvequadratic ():
    a = int(entry1.get())
    b = int(entry2.get())
    c = int(entry3.get())

    label1 = tk.Label(root, text= ((-b-cmath.sqrt((b**2)-(4*a*c))/(2*a))))
    canvas1.create_window(200, 230, window=label1)
    label1 = tk.Label(root, text = ((-b+cmath.sqrt((b**2)-(4*a*c))/(2*a))))
    canvas1.create_window(200, 260, window=label1)
    
button1 = tk.Button(text='Solve Quadratic', command=solvequadratic)
canvas1.create_window(200, 180, window=button1)

root.mainloop()

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Contributed on Feb 21 2022

DifficultBozo

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quadratic equation

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Δ=b2−4ac 
(-b±√Δ)/2a

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Contributed on Jan 30 2022

BGOPC

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Solving quadratic equations

Comment

import math

def solve_quadratic(a, b, c):
    # Calculate the discriminant
    discriminant = b**2 - 4*a*c
    
    # Check if the equation has complex roots
    if discriminant < 0:
        return "Complex Roots"
    
    # Calculate the roots
    root1 = (-b + math.sqrt(discriminant)) / (2*a)
    root2 = (-b - math.sqrt(discriminant)) / (2*a)
    
    # Return the roots
    return root1, root2

# Example usage
a = 1
b = -5
c = 6
roots = solve_quadratic(a, b, c)
print("Roots:", roots)

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Contributed on Sep 11 2023

YOU.com

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quadratic equation solver

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import cmath

def solve_quadratic(a, b, c):
    # Calculate the discriminant
    discriminant = (b**2) - (4*a*c)

    # Find two solutions
    solution1 = (-b - cmath.sqrt(discriminant)) / (2*a)
    solution2 = (-b + cmath.sqrt(discriminant)) / (2*a)

    return solution1, solution2

# Example usage
a = 1
b = -5
c = 6
solution1, solution2 = solve_quadratic(a, b, c)

print("Solutions: {0} and {1}".format(solution1, solution2))

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Contributed on Sep 12 2023

YOU.com

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quadratic formula

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from cmath import sqrt as c_sqrt
from math import sqrt as m_sqrt
import tkinter as tk


# creates and runs an instance of the window
window = tk.Tk()

#makes the size of the widow
window.geometry("600x400")

#sets a title for the screen
window.title("intercode")

# creating a new label and adding a text to the window
new_label=tk.Label(window,text='enter a,b,c quadratic formula\n')
new_label.pack()

#creates a entrybox to add words to window
entry_label1=tk.Entry(window)
entry_label1.pack()

entry_label2=tk.Entry(window)
entry_label2.pack()

entry_label3=tk.Entry(window)
entry_label3.pack()
# defines results as string var
results= tk.StringVar(window)



#creating a results label
result_lable=tk.Label(window,textvariable=results)
result_lable.pack()



def callbackfunct():
    a =entry_label1.get()
    b =entry_label2.get()
    c =entry_label3.get()
    try:
        a= float(a)
        b= float(b)
        c= float(c)

    except ValueError:
        results.set("\nEnter number.")


    if (b**2-4*a*c) < 0:
        minus = (-b-c_sqrt(b**2-4*a*c))/(2*a)
        plus = (-b+c_sqrt(b**2-4*a*c))/(2*a)
    else:
        minus = (-b-m_sqrt(b**2-4*a*c))/(2*a)
        plus = (-b+m_sqrt(b**2-4*a*c))/(2*a)
        
    # deletes the text in the text box
    entry_label1.delete(0,tk.END)
    entry_label2.delete(0,tk.END)
    entry_label3.delete(0,tk.END)

    var = (f"'B+'   {plus} , 'B-'   {minus}")
    results.set(var)
# creating a button with a command
buttonez = tk.Button(text="Submit", command=callbackfunct)
buttonez.pack()


window.mainloop()

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Contributed on Nov 03 2022

An7wan

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quadratic equation formula

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