solve using the quadratic formula. Round your answers to the nearest thousandth as needed. $$ 9 x^{2}=6 x-9 $$

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1. Write the given equation in standard form:
   $$
   9x^2 = 6x - 9 \quad \Rightarrow \quad 9x^2 - 6x + 9 = 0
   $$

2. Identify the coefficients in the standard quadratic form $$ ax^2 + bx + c = 0 $$:
   $$
   a = 9,\quad b = -6,\quad c = 9
   $$

3. Apply the quadratic formula:
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   Substitute the coefficients:
   $$
   x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(9)(9)}}{2(9)} = \frac{6 \pm \sqrt{36 - 324}}{18}
   $$

4. Simplify the expression under the square root:
   $$
   36 - 324 = -288 \quad \Rightarrow \quad x = \frac{6 \pm \sqrt{-288}}{18}
   $$

5. Express the square root of the negative number in terms of $$ i $$:
   $$
   \sqrt{-288} = i\sqrt{288}
   $$
   Notice that:
   $$
   \sqrt{288} = \sqrt{144 \times 2} = 12\sqrt{2}
   $$
   Thus:
   $$
   \sqrt{-288} = 12i\sqrt{2}
   $$

6. Substitute back into the formula:
   $$
   x = \frac{6 \pm 12i\sqrt{2}}{18}
   $$
   Simplify by dividing numerator and denominator by $$ 6 $$:
   $$
   x = \frac{6(1 \pm 2i\sqrt{2})}{18} = \frac{1 \pm 2i\sqrt{2}}{3}
   $$

7. For rounding, convert the real and imaginary parts to decimals:
   - The real part: 
     $$
     \frac{1}{3} \approx 0.333
     $$
   - The imaginary part: 
     $$
     \frac{2\sqrt{2}}{3} \approx \frac{2(1.414)}{3} \approx \frac{2.828}{3} \approx 0.943
     $$

8. Thus, the solutions are:
   $$
   x \approx 0.333 \pm 0.943i
   $$
The solutions are approximately $$ x \approx 0.333 + 0.943i $$ and $$ x \approx 0.333 - 0.943i $$.

solve using the quadratic formula. Round your answers to the nearest thousandth as needed. \( 9 x^{2}=6 x-9 \)

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