Solve the equation. \[ 9 x^{2}+21 x-8=0 \]

To solve the quadratic equation \(9x^2 + 21x - 8 = 0\), we can use the quadratic formula given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = 21\), and \(c = -8\). First, we calculate the discriminant: \[ b^2 - 4ac = 21^2 - 4 \cdot 9 \cdot (-8) = 441 + 288 = 729. \] Now, we take the square root of the discriminant: \[ \sqrt{729} = 27. \] Next, we apply the quadratic formula: \[ x = \frac{-21 \pm 27}{2 \cdot 9} = \frac{-21 \pm 27}{18}. \] Calculating the two possible values for \(x\): 1. For the positive root: \[ x = \frac{6}{18} = \frac{1}{3}. \] 2. For the negative root: \[ x = \frac{-48}{18} = -\frac{8}{3}. \] Thus, the solutions to the equation \(9x^2 + 21x - 8 = 0\) are: \[ x = \frac{1}{3} \quad \text{and} \quad x = -\frac{8}{3}. \]