Create a question about finding roots for a challenging quadratic equation by expanding and then factoring. The equation has brackets on both sides. One side can have a constant and/or a bracket and the other side can be a perfect square. The value of "a" cannot be equal to 1. Show the solution. The solution must be factorable - whole numbers. The question should have numbers in the teens.

### Question:
Solve the quadratic equation by expanding and then factoring:

\[
3(2x - 5)(x + 4) = (x - 7)^2
\]

### Solution:

1. **Expand both sides:**

   Start by expanding the left-hand side:
   \[
   3(2x - 5)(x + 4)
   \]
   First, apply the distributive property to the terms inside the parentheses:
   \[
   3 \left[ 2x(x + 4) - 5(x + 4) \right] = 3 \left[ 2x^2 + 8x - 5x - 20 \right]
   \]
   Simplifying inside the brackets:
   \[
   3(2x^2 + 3x - 20) = 6x^2 + 9x - 60
   \]

   Now expand the right-hand side:
   \[
   (x - 7)^2 = x^2 - 14x + 49
   \]

2. **Move all terms to one side of the equation:**

   Now, set the equation equal to zero by moving all terms to one side:
   \[
   6x^2 + 9x - 60 - (x^2 - 14x + 49) = 0
   \]
   Simplify by distributing the negative sign:
   \[
   6x^2 + 9x - 60 - x^2 + 14x - 49 = 0
   \]
   Combine like terms:
   \[
   (6x^2 - x^2) + (9x + 14x) + (-60 - 49) = 0
   \]
   \[
   5x^2 + 23x - 109 = 0
   \]

3. **Factor the quadratic:**

   We need to factor \(5x^2 + 23x - 109\). Start by finding two numbers whose product is \(5 \times (-109) = -545\) and whose sum is 23.

   These numbers are 34 and -16, since \(34 \times (-16) = -545\) and \(34 + (-16) = 23\).

   Rewrite the middle term using these numbers:
   \[
   5x^2 + 34x - 16x - 109 = 0
   \]
   Factor by grouping:
   \[
   (5x^2 + 34x) - (16x + 109) = 0
   \]
   Factor out common terms from each group:
   \[
   5x(x + 7) - 16(x + 7) = 0
   \]
   Factor out the common binomial factor:
   \[
   (5x - 16)(x + 7) = 0
   \]

4. **Solve for the roots:**

   Set each factor equal to zero:
   \[
   5x - 16 = 0 \quad \text{or} \quad x + 7 = 0
   \]
   Solving these:
   \[
   x = \frac{16}{5} \quad \text{or} \quad x = -7
   \]

### Final Answer:
The solutions to the quadratic equation are:
\[
x = \frac{16}{5} \quad \text{or} \quad x = -7
\]

---

Would you like further details on any part of this process or have any questions? Here are some follow-up questions to deepen your understanding:

1. What happens when the coefficient of \(x^2\) is 1 in a similar equation?
2. How would you solve the equation if both sides were perfect squares?
3. What is the geometric significance of the roots in this equation?
4. Can you always factor a quadratic equation with integer roots?
5. How would you solve this equation if it wasn't factorable using integers?

**Tip:** When factoring quadratics, always check the discriminant first. It tells you if real roots exist!

Create a question about finding roots for a challenging quadratic equation by expanding and then factoring. The equation has brackets on both sides. One side can have a constant and/or a bracket and the other side can be a perfect square. The value of 'a' cannot be equal to 1. Show the solution. The solution must be factorable - whole numbers. The question should have numbers in the teens.

Learn how to solve a challenging quadratic equation with numbers in the teens by expanding and factoring. The equation includes brackets on both sides, perfect squares, and requires factoring whole number solutions. A detailed step-by-step solution is provided, suitable for high school students.

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