一元二次方程的一般形式为 `ax^2 + bx + c = 0`，其中 `a`、`b`、`c` 为已知实数，且 `a` 不为 0。求解该方程可以使用求根公式：
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Python
```python
import math
def solve_quadratic(a, b, c):
discriminant = b 2 - 4 * a * c
if discriminant > 0:
root1 = (-b + math.sqrt(discriminant)) / (2 * a)
root2 = (-b - math.sqrt(discriminant)) / (2 * a)
return root1, root2
elif discriminant == 0:
root = -b / (2 * a)
return root, root
else:
realPart = -b / (2 * a)
imaginaryPart = math.sqrt(-discriminant) / (2 * a)
return complex(realPart, imaginaryPart), complex(realPart, -imaginaryPart)
示例方程: x^2 + 2x - 3 = 0
a = 1
b = 2
c = -3
solution = solve_quadratic(a, b, c)
print(f"The solutions are: {solution}")
```
Java
```java
import java.util.Scanner;
public class QuadraticEquationSolver {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.println("请输入a, b, c的值:");
double a = scanner.nextDouble();
double b = scanner.nextDouble();
double c = scanner.nextDouble();
double discriminant = b * b - 4 * a * c;
if (discriminant > 0) {
double root1 = (-b + Math.sqrt(discriminant)) / (2 * a);
double root2 = (-b - Math.sqrt(discriminant)) / (2 * a);
System.out.println("方程有两个不相等的实数根: " + root1 + " 和 " + root2);
} else if (discriminant == 0) {
double root = -b / (2 * a);
System.out.println("方程有两个相等的实数根: " + root);
} else {
double realPart = -b / (2 * a);
double imaginaryPart = Math.sqrt(-discriminant) / (2 * a);
System.out.println("方程有两个虚数根: " + realPart + " + " + imaginaryPart + "i 和 " + realPart + " - " + imaginaryPart + "i");
}
}
}
```
C++