Round #997

State the quadratic formula for solving ax^2 + bx + c = 0 and show how to use it to solve 2x^2 + 5x - 3 = 0.

```csharp using System; class Program { static void Main() { // define the coefficients double a = 2; double b = 5; double c = -3; // calculate the discriminant double discriminant = b * b - 4 * a * c; // calculate the two roots double root1 = (-b + Math.Sqrt(discriminant)) / (2 * a); double root2 = (-b - Math.Sqrt(discriminant)) / (2 * a); Console.WriteLine("The roots of the equation are: " + root1 + " and " + root2); } } ```

The quadratic formula for solving ax^2 + bx + c = 0 is x = (-b ± √(b²-4ac))/(2a). To solve 2x^2 + 5x - 3 = 0 using this formula, we first identify the coefficients: a = 2, b = 5, and c = -3. Then, we calculate the discriminant: b² - 4ac = 5² - 4*2*(-3) = 25 + 24 = 49. Next, we plug these values into the quadratic formula: x = (-5 ± √49)/(2*2) x = (-5 ± 7)/4 This gives us two roots: x = (-5 + 7)/4 = 2/4 = 1/2 x = (-5 - 7)/4 = -12/4 = -3 So, the solutions to the equation 2x^2 + 5x - 3 = 0 are x = 1/2 and x = -3. Here is a C# implementation of the quadratic formula: ```csharp using System; public class QuadraticSolver { public static (double, double) SolveQuadratic(double a, double b, double c) { double discriminant = b * b - 4 * a * c; if (discriminant < 0) { throw new InvalidOperationException("The equation has no real solutions."); } double sqrtDiscriminant = Math.Sqrt(discriminant); double root1 = (-b + sqrtDiscriminant) / (2 * a); double root2 = (-b - sqrtDiscriminant) / (2 * a); return (root1, root2); } public static void Main(string[] args) { double a = 2; double b = 5; double c = -3; try { var (root1, root2) = SolveQuadratic(a, b, c); Console.WriteLine($"The solutions to the equation {a}x^2 + {b}x + {c} = 0 are x = {root1} and x = {root2}."); } catch (InvalidOperationException ex) { Console.WriteLine(ex.Message); } } } ```