h) \( 3 y^{2}+10 y-48=0 \)

To solve the quadratic equation \( 3y^{2}+10y-48=0 \), we can use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 3 \), \( b = 10 \), and \( c = -48 \). Plugging in these values, we first find the discriminant: \[ b^2 - 4ac = 10^2 - 4 \cdot 3 \cdot (-48) = 100 + 576 = 676 \] Then we calculate \( y \): \[ y = \frac{-10 \pm \sqrt{676}}{2 \cdot 3} = \frac{-10 \pm 26}{6} \] This gives us two possible solutions: 1. \( y = \frac{16}{6} = \frac{8}{3} \) 2. \( y = \frac{-36}{6} = -6 \) So, the solutions to the equation \( 3y^{2}+10y-48=0 \) are \( y = \frac{8}{3} \) and \( y = -6 \).