Solve the equation. Express numbers in exact simplest form. \[ 4 q^{3}+21 q^{2}+20 q=0 \] The solution set is \( \{\square\} \).

To solve the equation \( 4q^{3} + 21q^{2} + 20q = 0 \), we first factor out the greatest common factor, which is \( q \): \[ q(4q^{2} + 21q + 20) = 0 \] This gives one solution: \( q = 0 \). Next, we need to solve the quadratic \( 4q^{2} + 21q + 20 = 0 \) using the quadratic formula: \[ q = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \] Here, \( a = 4 \), \( b = 21 \), and \( c = 20 \): \[ q = \frac{-21 \pm \sqrt{21^{2} - 4 \cdot 4 \cdot 20}}{2 \cdot 4} \] \[ = \frac{-21 \pm \sqrt{441 - 320}}{8} \] \[ = \frac{-21 \pm \sqrt{121}}{8} \] \[ = \frac{-21 \pm 11}{8} \] Calculating the two potential solutions: 1. \( q = \frac{-21 + 11}{8} = \frac{-10}{8} = \frac{-5}{4} \) 2. \( q = \frac{-21 - 11}{8} = \frac{-32}{8} = -4 \) Therefore, the complete solution set is: \[ \{ 0, -4, -\frac{5}{4} \} \]