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, we identify the coefficients: \( a = \frac{2}{3} \), \( b = 1 \), and \( c = \frac{1}{3} \). The quadratic formula is given by: \[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] First, calculate the discriminant (\( b^2 - 4ac \)): \[ b^2 = (1)^2 = 1 \] \[ 4ac = 4 \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{8}{9} \] \[ b^2 - 4ac = 1 - \frac{8}{9} = \frac{1}{9} \] Now, substitute back into the quadratic formula: \[ h = \frac{-1 \pm \sqrt{\frac{1}{9}}}{2 \cdot \frac{2}{3}} \] \[ h = \frac{-1 \pm \frac{1}{3}}{\frac{4}{3}} = \frac{-1 \pm \frac{1}{3}}{\frac{4}{3}} = \frac{-3 \pm 1}{4} \] This gives us two potential solutions: \[ h_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \] \[ h_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \] So, the solutions to the equation are: \[ \boxed{-\frac{1}{2}, -1} \]