Score: $2 / 5$ Penalty: none Question Show Examples A rocket is launched from a tower. The height of the rocket, $\mathrm{y}$ in feet, is related to the time after launch, $\mathrm{x}$ in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 1ooth of second. \[ y=-16 x^{2}+194 x+149 \] Answer Attempt 1 out of 2

Step 1 ：The rocket hits the ground when its height y is zero. So, we need to solve the equation \(-16x^2 + 194x + 149 = 0\) for x. This is a quadratic equation, and we can solve it using the quadratic formula: \(x = [-b ± sqrt(b^2 - 4ac)] / (2a)\) where a = -16, b = 194, and c = 149. We will get two solutions, but we only consider the positive one because time cannot be negative.

Step 2 ：Substitute a = -16, b = 194, and c = 149 into the quadratic formula. We get two solutions: \(x1 = -0.7247237512844675\) and \(x2 = 12.849723751284468\).

Step 3 ：Since time cannot be negative, we discard the negative solution. So, the time that the rocket will hit the ground is \(x2 = 12.849723751284468\) seconds.

Step 4 ：Rounding to the nearest hundredth of a second, the final answer is \(\boxed{12.85}\) seconds.