To solve a quadratic equation of the form `ax^2 + bx + c = 0`

, where `a`

, `b`

, and `c`

are constants and `x`

is the variable, you can use the quadratic formula:

`x = (-b ± sqrt(b^2 - 4ac)) / 2a`

Here is an example of how to solve a quadratic equation using Python:

```
import math
def solve_quadratic(a, b, c):
discriminant = b**2 - 4*a*c
if discriminant < 0:
return None # no real solutions
elif discriminant == 0:
return -b / (2*a) # one real solution
else:
root1 = (-b + math.sqrt(discriminant)) / (2*a)
root2 = (-b - math.sqrt(discriminant)) / (2*a)
return (root1, root2) # two real solutions
# test the function with a sample input
a = 1
b = -5
c = 6
print(solve_quadratic(a, b, c))
```

In this solution, we define a function `solve_quadratic`

that takes the coefficients `a`

, `b`

, and `c`

of a quadratic equation as input. The function first calculates the discriminant `b^2 - 4ac`

and checks whether it is negative (no real solutions), zero (one real solution), or positive (two real solutions).

If the discriminant is negative, the function returns `None`

to indicate that there are no real solutions. If the discriminant is zero, the function returns the single real solution `-b / 2a`

. If the discriminant is positive, the function calculates the two real solutions using the quadratic formula and returns them as a tuple.

In the main part of the script, we test the function with a sample input coefficients `a`

, `b`

, and `c`

and print out the result.

To use this code with other inputs, simply replace `a`

, `b`

, and `c`

with the desired coefficients. Note that this solution assumes that the input coefficients are valid and non-zero. You may want to add additional error checking to handle these cases.

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