Solve the quadratic equation by completing the square. \[ 3 z^{2}+2 z-1=0 \]

To solve the quadratic equation \( 3z^{2}+2z-1=0 \) by completing the square, we first divide the whole equation by 3 to simplify it: \[ z^{2} + \frac{2}{3}z - \frac{1}{3} = 0. \] Next, we move the constant term to the other side: \[ z^{2} + \frac{2}{3}z = \frac{1}{3}. \] Now, we need to complete the square on the left side. We take half of the coefficient of \( z \) (which is \( \frac{2}{3} \)), square it, and add it to both sides. Half of \( \frac{2}{3} \) is \( \frac{1}{3} \), and squaring it gives us \( \left(\frac{1}{3}\right)^2 = \frac{1}{9} \). Adding \( \frac{1}{9} \) to both sides results in: \[ z^{2} + \frac{2}{3}z + \frac{1}{9} = \frac{1}{3} + \frac{1}{9}. \] Now, we need a common denominator to combine the right side. The common denominator is 9, so: \[ \frac{1}{3} = \frac{3}{9} \implies \frac{3}{9} + \frac{1}{9} = \frac{4}{9}. \] Thus, we have: \[ \left(z + \frac{1}{3}\right)^{2} = \frac{4}{9}. \] Taking the square root of both sides gives: \[ z + \frac{1}{3} = \pm \frac{2}{3}. \] Now, we solve for \( z \): 1. For the positive case: \[ z + \frac{1}{3} = \frac{2}{3} \implies z = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}. \] 2. For the negative case: \[ z + \frac{1}{3} = -\frac{2}{3} \implies z = -\frac{2}{3} - \frac{1}{3} = -1. \] Thus, the solutions to the quadratic equation are: \[ z = \frac{1}{3} \quad \text{and} \quad z = -1. \]