Solve by factoring. (Enter your answers as a comma-separated list.) \[ 3 t^{2}-28 t=-9 \] \( t=\square \)

First, let's rearrange the equation to set it to zero: \[ 3t^2 - 28t + 9 = 0 \] Next, we need to factor the quadratic equation. To factor this, we look for two numbers that multiply to \(3 \cdot 9 = 27\) and add to \(-28\). The numbers are -27 and -1. Rewriting the middle term, we have: \[ 3t^2 - 27t - 1t + 9 = 0 \] Now, we factor by grouping: \[ 3t(t - 9) - 1(t - 9) = 0 \] This gives us: \[ (3t - 1)(t - 9) = 0 \] Setting each factor to zero gives us the solutions: 1. \( 3t - 1 = 0 \) which leads to \( t = \frac{1}{3} \) 2. \( t - 9 = 0 \) which leads to \( t = 9 \) Thus, the solutions are: \( t = \frac{1}{3}, 9 \)