To solve the quadratic equation (x2-11x+28=0), we’ll first understand its components and then apply the quadratic formula for its solution. This equation is in the standard form of a quadratic equation, (ax^2 + bx + c = 0), where (a), (b), and (c) are constants, and (x) represents an unknown variable. In our equation:

- (a = 1) (the coefficient of (x^2)),
- (b = -11) (the coefficient of (x)), and
- (c = 28) (the constant term).

The solutions to a quadratic equation can be found using the quadratic formula:

[x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}]

This formula computes the values of (x) that make the equation true, and it can yield two solutions, as indicated by the (\pm) symbol. These solutions are the points where the parabola, represented by the quadratic equation, crosses the x-axis.

Let’s break down the solution step by step:

**Calculate the discriminant**: The discriminant in a quadratic equation is given by (D = b^2 – 4ac). It helps determine the nature of the roots of the equation (whether they are real or complex, and whether they are distinct or equal).**Apply the quadratic formula**: Substitute the values of (a), (b), and (c) into the quadratic formula to find the values of (x).**Simplify**: After substitution, simplify the expression to find the roots of the equation.

Now, let’s solve the given equation (x^2 – 11x + 28 = 0) using these steps.

The discriminant (D = 9), which is positive, indicating that the equation has two distinct real roots.

By applying the quadratic formula and simplifying, we find the roots of the equation (x^2 – 11x + 28 = 0) to be:

- (x_1 = 7)
- (x_2 = 4)

These roots mean that the equation is satisfied (equals zero) when (x) is either 7 or 4. Graphically, this represents the points where the parabola (y = x^2 – 11x + 28) intersects the x-axis at (x = 7) and (x = 4).

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