Solve the inequality \( 3 y^{2}+4 y-32>0 \) Show your working clearly.

To solve the inequality \( 3y^2 + 4y - 32 > 0 \), we will first find the roots of the corresponding equation \( 3y^2 + 4y - 32 = 0 \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = 4 \), and \( c = -32 \). Plugging in these values: \[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-32)}}{2 \cdot 3} \] This simplifies to: \[ y = \frac{-4 \pm \sqrt{16 + 384}}{6} \] \[ y = \frac{-4 \pm \sqrt{400}}{6} \] Calculating \( \sqrt{400} = 20 \), we have: \[ y = \frac{-4 \pm 20}{6} \] Calculating the two possible roots: 1. \( y = \frac{-4 + 20}{6} = \frac{16}{6} = \frac{8}{3} \) 2. \( y = \frac{-4 - 20}{6} = \frac{-24}{6} = -4 \) Now that we have the roots, \( y = -4 \) and \( y = \frac{8}{3} \), we can analyze the sign of \( 3y^2 + 4y - 32 \) in the intervals defined by these roots: \( (-\infty, -4) \), \( (-4, \frac{8}{3}) \), and \( (\frac{8}{3}, +\infty) \). We'll test a value from each interval: 1. For the interval \( (-\infty, -4) \), let's choose \( y = -5 \): \[ 3(-5)^2 + 4(-5) - 32 = 75 - 20 - 32 = 23 > 0 \] 2. For the interval \( (-4, \frac{8}{3}) \), let's choose \( y = 0 \): \[ 3(0)^2 + 4(0) - 32 = -32 < 0 \] 3. For the interval \( (\frac{8}{3}, +\infty) \), let's choose \( y = 3 \): \[ 3(3)^2 + 4(3) - 32 = 27 + 12 - 32 = 7 > 0 \] Now we summarize the results: - The expression \( 3y^2 + 4y - 32 > 0 \) in the intervals \( (-\infty, -4) \) and \( (\frac{8}{3}, +\infty) \). In conclusion, the solution set to the inequality \( 3y^2 + 4y - 32 > 0 \) is: \[ y \in (-\infty, -4) \cup \left( \frac{8}{3}, +\infty \right) \]