https://file.aiquickdraw.com/mathbot/file/1729646491906xcl7bt9l.jpg

The quadratic equation shown in the image is:

\[
3z^2 + 8z + 3 = 6
\]

### Step 1: Simplify the equation
First, subtract 6 from both sides to set the equation equal to zero:

\[
3z^2 + 8z + 3 - 6 = 0 \quad \Rightarrow \quad 3z^2 + 8z - 3 = 0
\]

### Step 2: Factor the quadratic expression
We need to factor the quadratic \( 3z^2 + 8z - 3 \).

To factor, we look for two numbers that multiply to \( 3 \times -3 = -9 \) and add up to \( 8 \). These numbers are \( 9 \) and \( -1 \).

Now, split the middle term using these two numbers:

\[
3z^2 + 9z - z - 3 = 0
\]

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

\[
(3z^2 + 9z) - (z + 3) = 0
\]

Factor out the greatest common factor from each group:

\[
3z(z + 3) - 1(z + 3) = 0
\]

Now, factor out the common binomial factor \( (z + 3) \):

\[
(3z - 1)(z + 3) = 0
\]

### Step 4: Solve for \( z \)
Set each factor equal to zero:

1. \( 3z - 1 = 0 \quad \Rightarrow \quad z = \frac{1}{3} \)
2. \( z + 3 = 0 \quad \Rightarrow \quad z = -3 \)

### Final Solution:
The solutions are:

\[
z = \frac{1}{3}, \quad z = -3
\]

Let me know if you need more details or have any questions.

---

Here are 5 related questions to consider:

1. How would you factor a quadratic equation if it cannot be factored easily?
2. What is the quadratic formula, and when is it useful?
3. How can completing the square be used to solve quadratic equations?
4. What are the advantages of factoring vs. using the quadratic formula?
5. How can you determine if a quadratic equation has real solutions?

**Tip:** Always check your solutions by substituting them back into the original equation to ensure they satisfy it!

Use factoring techniques to solve the following quadratic equation: 3z^2 + 8z + 3 = 6.

Learn how to factor and solve the quadratic equation 3z^2 + 8z - 3 = 0 using step-by-step factoring techniques. This algebra problem demonstrates factoring by grouping and applying the zero product property. Suitable for students in Grades 9-11.

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