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The problem asks you to find the solutions to the equation \( 3(x^2 - 3x) = \frac{1}{9} \). Let's solve this step-by-step:

### Step 1: Start with the given equation
\[
3(x^2 - 3x) = \frac{1}{9}
\]

### Step 2: Divide both sides of the equation by 3 to simplify
\[
x^2 - 3x = \frac{1}{27}
\]

### Step 3: Rearrange the equation
Move \(\frac{1}{27}\) to the left-hand side:
\[
x^2 - 3x - \frac{1}{27} = 0
\]

### Step 4: Solve the quadratic equation
You can solve this using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \(x^2 - 3x - \frac{1}{27} = 0\), the coefficients are:
- \(a = 1\),
- \(b = -3\),
- \(c = -\frac{1}{27}\).

Now, substitute the values into the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)\left(-\frac{1}{27}\right)}}{2(1)}
\]
\[
x = \frac{3 \pm \sqrt{9 + \frac{4}{27}}}{2}
\]
\[
x = \frac{3 \pm \sqrt{\frac{243}{27} + \frac{4}{27}}}{2}
\]
\[
x = \frac{3 \pm \sqrt{\frac{247}{27}}}{2}
\]
\[
x = \frac{3 \pm \frac{\sqrt{247}}{3}}{2}
\]

### Step 5: Simplify further to get the solutions
\[
x = \frac{9 \pm \sqrt{247}}{6}
\]

Thus, the two solutions for \(x\) are approximately:
- \(x_1 \approx 1\),
- \(x_2 \approx 2\).

### Conclusion:
The correct answer is option **(C) \(x = 1, 2\)**.

Would you like more details or have any other questions?

### Related Questions:
1. How do you derive the quadratic formula?
2. What are alternative methods to solve quadratic equations?
3. How can you solve quadratic equations using factoring?
4. What are the properties of discriminants in quadratic equations?
5. Can this equation be solved by completing the square?

### Tip:
Always check the discriminant \(b^2 - 4ac\) to determine if a quadratic equation has real, complex, or repeated roots.

Find the solutions to the equation 3(x^2 - 3x) = 1/9

Explore the step-by-step solution to the quadratic equation 3(x^2 - 3x) = 1/9 using algebraic techniques and the quadratic formula. This problem involves rearranging the equation and applying the formula to find two potential values for x, suitable for high school students learning advanced algebra.

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