# Find the solutions to this system y=x^2-2x-2

Find the solutions to this system y=x^2-2x-2. We share this issue with you in detail. You can view all the answers below.

## Find the solutions to this system y=x^2-2x-2

Certainly! Here’s a detailed article explaining how to find the solutions to the equation (y = x^2 – 2x – 2).

• Solving the Quadratic Equation (y = x^2 – 2x – 2)

A quadratic equation is a second-degree polynomial equation that can be written in the form (ax^2 + bx + c = 0), where (a), (b), and (c) are constants. In this case, the given equation (y = x^2 – 2x – 2) is already in the form (y = ax^2 + bx + c), with (a = 1), (b = -2), and (c = -2).

To solve for (x), we need to find the values of (x) that satisfy the equation. This involves using methods like factoring, completing the square, or using the quadratic formula.

### Method 1: Factoring

Find the solutions to this system y=x^2-2x-2:

1. Start with the equation: (y = x^2 – 2x – 2).
2. Set (y) equal to zero to form the quadratic equation: (x^2 – 2x – 2 = 0).
3. Factor the quadratic expression: ((x – \alpha)(x – \beta) = 0), where (\alpha) and (\beta) are the solutions.
4. Expand the factored form: (x^2 – (\alpha + \beta)x + \alpha\beta = 0).
5. Compare the coefficients to find (\alpha + \beta = 2) and (\alpha\beta = -2).
6. Solve the system of equations to find (\alpha) and (\beta), which will be the solutions to the quadratic equation.

Find the solutions to this system y=x^2-2x-2:

1. Start with the equation: (y = x^2 – 2x – 2).
2. Set (y) equal to zero to form the quadratic equation: (x^2 – 2x – 2 = 0).
3. Use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}), where (a = 1), (b = -2), and (c = -2).
4. Substitute the values into the formula and calculate the two possible solutions for (x).

### Method 3: Completing the Square

Find the solutions to this system y=x^2-2x-2:

1. Start with the equation: (y = x^2 – 2x – 2).
2. Set (y) equal to zero to form the quadratic equation: (x^2 – 2x – 2 = 0).
3. Complete the square by adding and subtracting the square of half of the coefficient of (x): (x^2 – 2x + 1 – 2 – 1 = 0).
4. Factor the perfect square trinomial: ((x – 1)^2 – 3 = 0).
5. Solve for (x) by isolating ((x – 1)^2), then finding the square root of both sides.

#### Our Last Words

Find the solutions to this system y=x^2-2x-2 / Once you’ve used one of the methods to solve the quadratic equation (y = x^2 – 2x – 2), you’ll have found the values of (x) that satisfy the equation. These values will be the solutions to the equation. Remember that quadratic equations can have two real solutions, one real solution (if the discriminant is zero), or two complex solutions (if the discriminant is negative).

We also shared our last words on this subject for you. If you have any other questions, feel free to write them in the comments.