Question Use the square roots property to solve the quadratic equation \( 9 q^{2}+48 q+64=100 \). If there are multiple answers, list them separated by a comma, e.g. 1,2 . If there is no solution, enter \( \varnothing \).

To solve the equation \( 9 q^{2} + 48 q + 64 = 100 \), we first set the equation to zero: \[ 9 q^{2} + 48 q + 64 - 100 = 0 \] \[ 9 q^{2} + 48 q - 36 = 0 \] Next, we simplify by dividing the entire equation by 9: \[ q^{2} + \frac{48}{9} q - 4 = 0 \] \[ q^{2} + \frac{16}{3} q - 4 = 0 \] Now we can use the square root property. The quadratic formula can also be applied here, \[ q = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]: Here, \( a = 1 \), \( b = \frac{16}{3} \), and \( c = -4 \): Calculating \( b^2 - 4ac \): \[ \left(\frac{16}{3}\right)^2 - 4 \cdot 1 \cdot (-4) = \frac{256}{9} + 16 = \frac{256}{9} + \frac{144}{9} = \frac{400}{9} \] Now applying the quadratic formula: \[ q = \frac{-\frac{16}{3} \pm \sqrt{\frac{400}{9}}}{2} \] \[ q = \frac{-\frac{16}{3} \pm \frac{20}{3}}{2} \] Calculating the two possible values: 1. \[ q = \frac{-\frac{16}{3} + \frac{20}{3}}{2} = \frac{\frac{4}{3}}{2} = \frac{2}{3} \] 2. \[ q = \frac{-\frac{16}{3} - \frac{20}{3}}{2} = \frac{-\frac{36}{3}}{2} = \frac{-12}{2} = -6 \] So the solutions are: \( \frac{2}{3}, -6 \).