2n ^2 +12n−1n−6=0.    Use factoring form only

To solve the quadratic equation \( 2n^2 + 12n - 1n - 6 = 0 \) by factoring, let's follow these steps:

### Step 1: Combine like terms
First, combine the \(12n\) and \(-1n\) terms:

\[
2n^2 + 11n - 6 = 0
\]

### Step 2: Factor the quadratic expression
We need to find two numbers that multiply to \(2 \times (-6) = -12\) and add up to \(11\).

The numbers that work are \(12\) and \(-1\) because:

\[
12 \times (-1) = -12 \quad \text{and} \quad 12 + (-1) = 11
\]

Now, we rewrite the middle term \(11n\) using these two numbers:

\[
2n^2 + 12n - 1n - 6 = 0
\]

### Step 3: Group terms and factor by grouping
Group the terms into two pairs:

\[
(2n^2 + 12n) - (1n + 6) = 0
\]

Factor out the greatest common factor from each group:

\[
2n(n + 6) - 1(n + 6) = 0
\]

### Step 4: Factor out the common binomial
Now, factor out the common binomial factor \((n + 6)\):

\[
(n + 6)(2n - 1) = 0
\]

### Step 5: Solve for \(n\)
Set each factor equal to zero and solve for \(n\):

1. \( n + 6 = 0 \)

\[
n = -6
\]

2. \( 2n - 1 = 0 \)

\[
2n = 1 \quad \Rightarrow \quad n = \frac{1}{2}
\]

### Final Answer:
The solutions to the equation are:

\[
n = -6 \quad \text{or} \quad n = \frac{1}{2}
\]

Would you like further details or have any questions?

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Here are 5 related questions:
1. How can you verify that your factorization is correct?
2. What are some common mistakes to avoid when factoring quadratic equations?
3. Can you apply the factoring method to solve cubic equations?
4. How would the process change if the quadratic equation was not factorable over the integers?
5. How does factoring relate to the quadratic formula?

**Tip:** Always check your factored solution by expanding it to ensure it equals the original quadratic equation.

Learn how to solve the quadratic equation 2n^2 + 12n - n - 6 = 0 using the factoring method with a step-by-step solution. This problem involves algebraic concepts such as factoring quadratic equations. Suitable for students in Grades 9-12.

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