Once you set $h(t)$ equal to that value, you will have a quadratic equation in $t$. Can you put it into the standard form $at^2 + bt + c = 0$?

This quadratic does not factor nicely. Use the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with your values of $a$, $b$, and $c$.

You should get two values for $t$. Think about which value makes sense for time in this physical situation.

Type y = -5x^2 + 40x + 3 into Desmos. This graphs the height (y) of the firework as a function of time (x).

Zoom or pan so you can see where the parabola crosses the x-axis (where $y = 0$). Tap or click on the right-hand x-intercept; note the positive x-value shown. That x-value is the time in seconds when the firework returns to the ground.

The function $h(t)=-5t^2+40t+3$ gives the height (in meters) at time $t$ (in seconds).

"Return to the ground" means the height is $0$ meters. So we need to solve:

for $t$.

It is easier to use the quadratic formula if the coefficient of $t^2$ is positive. Multiply both sides by $-1$:

Now the equation is in standard form $at^2 + bt + c = 0$ with:

• $a = 5$
• $b = -40$
• $c = -3$.

The quadratic formula is

First compute the discriminant $b^2 - 4ac$:

Now plug into the formula:

Estimate $\sqrt{1660}$. It is a bit more than $40$; a good approximation is $\sqrt{1660} \approx 40.7$.

So the two solutions are approximately:

A negative time does not make sense in this context, so we discard $t \approx -0.07$.

Therefore, the model predicts the firework returns to the ground at about 8.07 seconds, which matches choice C) 8.07.