Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary. \[ 3 a^{2}+12 a<4(a+4) \] The solution set is \( \square \)

First, let's simplify the inequality step-by-step. Start with the original inequality: \[ 3a^2 + 12a < 4(a + 4) \] Expanding the right side gives: \[ 3a^2 + 12a < 4a + 16 \] Now, subtract \(4a\) and \(16\) from both sides: \[ 3a^2 + 12a - 4a - 16 < 0 \] This simplifies to: \[ 3a^2 + 8a - 16 < 0 \] Next, we can factor the quadratic expression. To find the roots, we can use the quadratic formula: \[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 3\), \(b = 8\), and \(c = -16\). Plugging in the values gives: \[ a = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot (-16)}}{2 \cdot 3} \] Calculating the discriminant: \[ 64 + 192 = 256 \] So: \[ \sqrt{256} = 16 \] Now, substituting back in: \[ a = \frac{-8 \pm 16}{6} \] This yields two solutions: 1. \(a = \frac{8}{6} = \frac{4}{3}\) 2. \(a = \frac{-24}{6} = -4\) Now we will test intervals created by these roots, \(-4\) and \(\frac{4}{3}\): 1. For \(a < -4\), say \(a = -5\): \[ 3(-5)^2 + 8(-5) - 16 = 75 - 40 - 16 = 19 > 0 \] 2. For \(-4 < a < \frac{4}{3}\), say \(a = 0\): \[ 3(0)^2 + 8(0) - 16 = -16 < 0 \] 3. For \(a > \frac{4}{3}\), say \(a = 2\): \[ 3(2)^2 + 8(2) - 16 = 12 + 16 - 16 = 12 > 0 \] From this testing, the inequality \(3a^2 + 8a - 16 < 0\) holds for the interval: \[ (-4, \frac{4}{3}) \] Thus, the solution set in interval notation is: \[ \boxed{(-4, \frac{4}{3})} \]