Find the real solutions, if any, of the following equation. Use the quadratic formula. \( \frac{2}{9} x^{2}-x=\frac{1}{9} \) Select the correct choice below and, if necessary, fill in the answer box to complete your answer. A. The solution set is \( \} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The solution is not a real number.

First, let's rearrange the equation to the standard quadratic form by bringing all terms to one side: \[ \frac{2}{9} x^{2} - x - \frac{1}{9} = 0 \] Now, we can utilize the quadratic formula, which is: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] Here, \(a = \frac{2}{9}\), \(b = -1\), and \(c = -\frac{1}{9}\). Plugging in these values, we first calculate the discriminant (\(b^2 - 4ac\)): \[ b^{2} = (-1)^{2} = 1 \] \[ 4ac = 4 \cdot \frac{2}{9} \cdot -\frac{1}{9} = -\frac{8}{81} \] Thus, the discriminant is: \[ 1 - \left(-\frac{8}{81}\right) = 1 + \frac{8}{81} = \frac{81}{81} + \frac{8}{81} = \frac{89}{81} \] Since the discriminant is positive (\(\frac{89}{81} > 0\)), we have two distinct real solutions. Now, substituting everything into our quadratic formula gives: \[ x = \frac{-(-1) \pm \sqrt{\frac{89}{81}}}{2 \cdot \frac{2}{9}} = \frac{1 \pm \frac{\sqrt{89}}{9}}{\frac{4}{9}} = \frac{9(1 \pm \sqrt{89})}{4} \] So the two solutions are: \[ x = \frac{9(1 + \sqrt{89})}{4}, \quad x = \frac{9(1 - \sqrt{89})}{4} \] Thus, your answer is: The solution set is \( \left\{ \frac{9(1 + \sqrt{89})}{4}, \frac{9(1 - \sqrt{89})}{4} \right\} \)