A rocket is launched from a tower. The height of the rocket, \( y \) in feet, is related to the time after launch, \( x \) 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. \[ y=-16 x^{2}+174 x+84 \]

We are given the height function \[ y = -16x^2 + 174x + 84 \] The rocket hits the ground when \( y = 0 \). So we solve: \[ -16x^2 + 174x + 84 = 0 \] **Step 1. Rewrite the Equation** Multiply both sides by \(-1\) to simplify the quadratic: \[ 16x^2 - 174x - 84 = 0 \] **Step 2. Use the Quadratic Formula** The quadratic formula for an equation \( ax^2 + bx + c = 0 \) is \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 16 \), \( b = -174 \), and \( c = -84 \). Substitute these into the formula: \[ x = \frac{-(-174) \pm \sqrt{(-174)^2 - 4(16)(-84)}}{2(16)} \] Simplify the numerator: \[ x = \frac{174 \pm \sqrt{174^2 - 4(16)(-84)}}{32} \] **Step 3. Calculate the Discriminant** Compute \( 174^2 \): \[ 174^2 = 30276 \] Compute \( -4(16)(-84) \) (note that the negatives cancel): \[ -4(16)(-84) = 4 \times 16 \times 84 = 5376 \] Now add these values: \[ \text{Discriminant} = 30276 + 5376 = 35652 \] Thus, we have: \[ x = \frac{174 \pm \sqrt{35652}}{32} \] **Step 4. Evaluate the Square Root** We approximate the square root: \[ \sqrt{35652} \approx 188.8 \] **Step 5. Find the Acceptable Root** Substitute the approximate value into the quadratic formula: \[ x = \frac{174 \pm 188.8}{32} \] We have two solutions: 1. \( x = \frac{174 + 188.8}{32} \approx \frac{362.8}{32} \approx 11.34 \) 2. \( x = \frac{174 - 188.8}{32} \approx \frac{-14.8}{32} \approx -0.46 \) Since time cannot be negative, we discard the negative solution. **Step 6. State the Final Answer** Rounding \( 11.34 \) to the nearest tenth of a second gives: \[ x \approx 11.3 \text{ seconds} \] Thus, the rocket will hit the ground after approximately \( 11.3 \) seconds.