We can solve the equation \((x+7)(x-3) = -7\) step by step.

### Step 1: Expand the left-hand side
We use the distributive property (FOIL method) to expand:
\[
(x+7)(x-3) = x^2 - 3x + 7x - 21 = x^2 + 4x - 21
\]
Now the equation becomes:
\[
x^2 + 4x - 21 = -7
\]

### Step 2: Move all terms to one side
Add 7 to both sides to set the equation equal to 0:
\[
x^2 + 4x - 21 + 7 = 0 \implies x^2 + 4x - 14 = 0
\]

### Step 3: Solve the quadratic equation
We now solve the quadratic equation \(x^2 + 4x - 14 = 0\) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 1\), \(b = 4\), and \(c = -14\). Plugging these values into the formula:
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-14)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 56}}{2} = \frac{-4 \pm \sqrt{72}}{2}
\]
Simplify \(\sqrt{72}\):
\[
\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}
\]
Now substitute this back into the equation:
\[
x = \frac{-4 \pm 6\sqrt{2}}{2}
\]
Simplify:
\[
x = -2 \pm 3\sqrt{2}
\]

### Final Answer:
The two solutions are:
\[
x = -2 + 3\sqrt{2} \quad \text{or} \quad x = -2 - 3\sqrt{2}
\]

Would you like more details or have any questions?

Here are some related questions:
1. How do you complete the square for solving quadratic equations?
2. What are the steps for using the quadratic formula in detail?
3. How does the discriminant help in determining the nature of the roots of a quadratic equation?
4. Can quadratic equations always be solved by factoring?
5. What is the relationship between quadratic equations and parabolas?

**Tip:** Always check the discriminant (\(b^2 - 4ac\)) to quickly determine if a quadratic equation has real, imaginary, or repeated roots.

Explore a detailed solution to the quadratic equation (x+7)(x-3) = -7. Learn how to expand the expression, use the quadratic formula, and apply the discriminant to find the roots. This step-by-step guide is perfect for students in Grades 8-10.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation