To solve the quadratic equation \(3x^2 + 6x + 2 = 0\) using the quadratic formula, follow these steps:

### Step 1: Identify the coefficients
The quadratic equation is in the form \(ax^2 + bx + c = 0\). For the equation \(3x^2 + 6x + 2 = 0\):
- \(a = 3\)
- \(b = 6\)
- \(c = 2\)

### Step 2: Write the quadratic formula
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

### Step 3: Substitute the values of \(a\), \(b\), and \(c\) into the formula
Substituting the given values:
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(3)(2)}}{2(3)}
\]

### Step 4: Simplify under the square root
First, calculate the discriminant (\(b^2 - 4ac\)):
\[
6^2 - 4(3)(2) = 36 - 24 = 12
\]
So,
\[
x = \frac{-6 \pm \sqrt{12}}{6}
\]

### Step 5: Simplify the square root
Simplify \(\sqrt{12}\):
\[
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
\]
So the equation becomes:
\[
x = \frac{-6 \pm 2\sqrt{3}}{6}
\]

### Step 6: Simplify the final expression
Factor out and simplify:
\[
x = \frac{-6}{6} \pm \frac{2\sqrt{3}}{6} = -1 \pm \frac{\sqrt{3}}{3}
\]

### Step 7: Write the final solutions
The solutions are:
\[
x_1 = -1 + \frac{\sqrt{3}}{3}, \quad x_2 = -1 - \frac{\sqrt{3}}{3}
\]

These are the two solutions in their simplified exact form.

Would you like any further explanations or do you have any questions?

Here are five related questions you might find useful:
1. How can you approximate the solutions of a quadratic equation to decimal places?
2. What is the significance of the discriminant being positive, zero, or negative?
3. How can you determine the axis of symmetry for a quadratic equation?
4. What are the steps to solve a quadratic equation by factoring?
5. How can the quadratic formula be derived from completing the square?

**Tip:** Always simplify the square root as much as possible to make your final solutions cleaner and more interpretable.

Learn how to solve the quadratic equation 3x² + 6x + 2 = 0 using the quadratic formula with a detailed step-by-step solution. This problem covers key algebraic concepts, including the quadratic formula and discriminant. Suitable for students in Grades 10-12.

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