Questions: Solve by applying the zero product property. 15t(t+1)=t-3 If there is more than one solution, separate the answers with commas. The solution set is

Solution Steps

First, we start with the given equation: \[ 15t(t + 1) = t - 3 \]

We rearrange it to set it equal to zero: \[ 15t(t + 1) - (t - 3) = 0 \]

Next, we simplify the left-hand side: \[ 15t^2 + 15t - t + 3 = 0 \] \[ 15t^2 + 14t + 3 = 0 \]

We solve the quadratic equation \(15t^2 + 14t + 3 = 0\) using the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 15\), \(b = 14\), and \(c = 3\).

Calculate the discriminant: \[ \Delta = b^2 - 4ac = 14^2 - 4 \cdot 15 \cdot 3 = 196 - 180 = 16 \]

Using the quadratic formula: \[ t = \frac{-14 \pm \sqrt{16}}{2 \cdot 15} = \frac{-14 \pm 4}{30} \]

This gives us two solutions: \[ t_1 = \frac{-14 + 4}{30} = \frac{-10}{30} = -\frac{1}{3} \] \[ t_2 = \frac{-14 - 4}{30} = \frac{-18}{30} = -\frac{3}{5} \]

Final Answer