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To solve for \( x \) in the equation \( 2x^2 - 7x - 30 = 0 \), we can try factoring it.

### Step 1: Setting up the equation
The equation is:
\[
2x^2 - 7x - 30 = 0
\]

### Step 2: Factoring
To factor, we look for two numbers that multiply to \( 2 \cdot (-30) = -60 \) and add up to \( -7 \).

The numbers that work are \( 5 \) and \( -12 \) because:
\[
5 \times (-12) = -60 \quad \text{and} \quad 5 + (-12) = -7
\]

### Step 3: Splitting the middle term
Rewrite \( -7x \) as \( 5x - 12x \):
\[
2x^2 + 5x - 12x - 30 = 0
\]

### Step 4: Factoring by grouping
Group the terms:
\[
(2x^2 + 5x) - (12x + 30) = 0
\]

Factor out \( x \) from the first group and \( -6 \) from the second group:
\[
x(2x + 5) - 6(2x + 5) = 0
\]

Now, factor out \( 2x + 5 \):
\[
(2x + 5)(x - 6) = 0
\]

### Step 5: Solving for \( x \)
Set each factor to zero:
1. \( 2x + 5 = 0 \Rightarrow x = -\frac{5}{2} \)
2. \( x - 6 = 0 \Rightarrow x = 6 \)

### Solution
The solutions are:
\[
x = -\frac{5}{2} \quad \text{or} \quad x = 6
\]

### Answer
The correct option is:
\[
\text{Option A: } x = -\frac{5}{2} \text{ or } x = 6
\]

Would you like further details or have any questions?

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Here are five related questions:
1. How can you verify the solutions by substituting back into the original equation?
2. Why does factoring by grouping work in this problem?
3. What are other methods to solve quadratic equations, and when are they useful?
4. How does the discriminant in a quadratic equation help determine the type of solutions?
5. How would the solution change if the equation was \( 2x^2 - 7x + 30 = 0 \)?

**Tip:** When factoring quadratics, always check if the quadratic is factorable before applying other methods like the quadratic formula.

Solve for x in the equation 2x^2 - 7x - 30 = 0.

Learn how to solve the quadratic equation 2x^2 - 7x - 30 = 0 by factoring. This solution provides a step-by-step approach, covering key algebraic concepts and using the Zero Product Property. Suitable for students in Grades 8-10.

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