Solve \( \frac{2}{3} h^{2}+h=-\frac{1}{3} \) by using the quadratic formula. Give an exact answer and simplify any fractions. If there are multiple answers, list them separated by a comma, e.g. 1,2 . If there is no solution, enter \( \varnothing \). Provide your answer below:

To solve the equation \( \frac{2}{3} h^{2}+h+\frac{1}{3} = 0 \) using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we identify the coefficients as follows: \( a = \frac{2}{3} \), \( b = 1 \), and \( c = \frac{1}{3} \). First, we calculate the discriminant \( b^2 - 4ac \): \[ b^2 = 1^2 = 1 \] \[ 4ac = 4 \times \frac{2}{3} \times \frac{1}{3} = \frac{8}{9} \] \[ b^2 - 4ac = 1 - \frac{8}{9} = \frac{1}{9} \] Now we can substitute into the quadratic formula: \[ h = \frac{-1 \pm \sqrt{\frac{1}{9}}}{2 \times \frac{2}{3}} = \frac{-1 \pm \frac{1}{3}}{\frac{4}{3}} = \frac{-3 \pm 1}{4} \] This gives us two potential solutions: 1. \( h = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \) 2. \( h = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \) Thus, the solution to the equation is \[ -\frac{1}{2}, -1 \]