input: Write a function to solve a quadratic equation in Python. The function should take three parameters, a, b, and c, which represent the coefficients of the quadratic equation a*x^2 + b*x + c = 0. If the equation has real roots, the function should return a dictionary with the status set to True and the roots as a list. If there are no real roots, the function should return a dictionary with the status set to False and an empty list for the roots. If there is only one real root, it should be returned in a list with one element.
  
  import math
  
  def solveQuadratic(a, b, c): 
      discriminant = (b**2) - (4*a*c);
      
      # If discriminant is negative, there are no real roots.
      if (discriminant < 0): 
          return {
              "status": False,
              "roots": []
          }
      # If discriminant is zero, there is only one real roots.
      elif (discriminant == 0):
          x = -b/(2*a)
          return {
              "status": True,
              "roots": [x]
          }
      # Otherwise, there are 2 real roots 
      else:
          x1 = (-b + math.sqrt(discriminant)) / (2*a); 
          x2 = (-b - math.sqrt(discriminant)) / (2*a); 
          return {
              "status": True,
              "roots": [x1, x2]
          }
  
  assert solveQuadratic(1, 3, -4) == {'status': True, 'roots': [-4.0, 1.0]}
  assert solveQuadratic(1, 2, 1) == {'status': True, 'roots': [-1.0]}
  assert solveQuadratic(1, 0, 1) == {'status': False, 'roots': []}
output: import math
  
  def solveQuadratic(a, b, c): 
      discriminant = (b**2) - (4*a*c)
      
      if discriminant < 0: 
          return {
              "status": False,
              "roots": []
          }
      elif discriminant == 0:
          x = -b/(2*a)
          return {
              "status": True,
              "roots": [x]
          }
      else:
          x1 = (-b + math.sqrt(discriminant)) / (2*a)
          x2 = (-b - math.sqrt(discriminant)) / (2*a)
          return {
              "status": True,
              "roots": [x1, x2]
          }