Implement function which will return sum of roots of a quadratic equation rounded to 2 decimal places, if there are any possible roots, else return **None/null/nil/nothing**. If you use discriminant,when discriminant = 0, x1 = x2 = root => return sum of both roots. There will always be valid arguments. 

Quadratic equation - https://en.wikipedia.org/wiki/Quadratic_equation

Difficulty: introductory

Starter code:

def roots(a,b,c):
	

Solutions:
def roots(a,b,c):
    if b**2>=4*a*c:
        return round(-b/a,2)

def roots(a,b,c):
    import math
    d = b ** 2 - 4 * a * c
    if d > 0:
        return round(-2 * b / (2 * a), 2)
    elif d == 0:
        x = -b / (2 * a)
        return x * 2
    else:
        return None

def roots(a,b,c):
    return round(-b / a, 2) if b * b - 4 * a * c >= 0 else None  


def roots(a,b,c):
    # your code
    d = b**2 - 4*a*c
    if d < 0:
        return None
    elif d == 0:
        return (-b + d**0.5)/(2*a) * 2
    else:
        x1 = (-b + d**0.5)/(2*a)
        x2 = (-b - d**0.5)/(2*a)
        return round(x1 + x2, 2)



def roots(a ,b, c):
    d = b**2 - 4*a*c
    if d >= 0:
        return round(-b / a, 2)

roots=lambda a,b,c:None if b*b<a*c*4else round(-b/a,2)

def roots(a,b,c):
    # your code
    D = b**2-4*a*c
    if D<0:
        #No solution in the set of real numbers
        return None
    if D==0:
        #Same "double" solution
        x= (-b+D**0.5)/(2*a)
        return round(x*2,2)
    if D>0:
        #Two different solutions
        x1 = (-b+D**0.5)/(2*a)
        x2 = (-b-D**0.5)/(2*a)
        return round((x1+x2),2)

import math

def roots(a,b,c):
    dis = b ** 2 - 4 * a * c
    if dis < 0: return None
    return round(((-b + math.sqrt(dis)) / (2*a)) + ((-b - math.sqrt(dis)) / (2*a)), 2)

roots=lambda a, b, c: round(-(b/(2*a))*2, 2) if (b**2)-(4*a*c) >= 0 else None

import math
def roots(a,b,c):
    d=round(b*b-4*a*c,2) 
    if d<0 :
        return None
    return round(-b/a,2)