The equation $y = -6t^2 - 10t + 56$ describes the height (in feet) of a ball thrown downward at 10 feet per second from a height of 56 feet from the surface from Mars. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.
Setting $y$ to zero, we find the following:
\begin{align*}
-6t^2 - 10t + 56 &= 0 \\
\Rightarrow \quad 6t^2 + 10t - 56 &= 0 \\
\Rightarrow \quad 3t^2 + 5t - 28 &= 0 \\
\Rightarrow \quad (3t-7)(t+4) &= 0.
\end{align*}As $t$ must be positive, we can see that $t = \frac{7}{3} \approx \boxed{2.33}.$
2.33
Level 4
Algebra
Problem: The equation $y = -6t^2 - 10t + 56$ describes the height (in feet) of a ball thrown downward at 10 feet per second from a height of 56 feet from the surface from Mars. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find when the ball hits the ground, set the height equation equal to zero: $-6t^2 - 10t + 56 = 0$. This is a quadratic equation in standard form $at^2 + bt + c = 0$, where $a = -6$, $b = -10$, and $c = 56$. 

Next, apply the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. First, calculate the discriminant: $b^2 - 4ac = (-10)^2 - 4(-6)(56) = 100 + 1344 = 1444$. 

Now, take the square root of the discriminant: $\sqrt{1444} = 38$. Substitute back into the quadratic formula: 

$t = \frac{-(-10) \pm 38}{2(-6)} = \frac{10 \pm 38}{-12}$. 

This gives two potential solutions: 

1. $t = \frac{48}{-12} = -4$ (not valid since time cannot be negative).
2. $t = \frac{-28}{-12} = \frac{28}{12} = \frac{7}{3} \approx 2.33$.

Thus, the time it takes for the ball to hit the ground is approximately 2.33 seconds.<|END_OF_LATENT|> The final answer is $2.33$.