After you set the equation equal to 80, move all terms to one side to get 0 on the other side. Can you divide by a common factor to make the numbers smaller?

Once you have a simplified quadratic like $t^2 - 3t + 1 = 0$, it does not factor nicely. What formula can you always use to solve any quadratic equation?

The quadratic gives you two times when the projectile is at 80 feet. Think about the shape of a projectile's path (up then down). Which of the two times must be the earlier one?

In Desmos, type h(t) = -16t^2 + 48t + 64 or, using x as the variable, y = -16x^2 + 48x + 64 so you can see the projectile's height as a parabola.

On a new line, type y = 80 to draw a horizontal line showing a height of 80 feet.

Look at the two intersection points of the parabola and the line y = 80. Click each point and note their x-coordinates; these are the two times when the projectile is at 80 feet.

Compare the two x-coordinates and identify the smaller positive value. That smaller x-value is the time when the projectile first reaches 80 feet.

We are told the height in feet is modeled by

We want the times when the height is 80 feet, so set $h(t) = 80$:

Move 80 to the left side to set the equation equal to 0:

which simplifies to

Divide every term by $-16$ to simplify:

Now we have a simpler quadratic equation in standard form $at^2+bt+c=0$.

For $t^2 - 3t + 1 = 0$, we have $a = 1$, $b = -3$, and $c = 1$.

The quadratic formula is

Substitute $a$, $b$, and $c$:

So there are two possible times when the height is 80 feet, given by these two expressions.

Both values $t = \dfrac{3 + \sqrt{5}}{2}$ and $t = \dfrac{3 - \sqrt{5}}{2}$ are positive, so they both correspond to real times when the projectile is at 80 feet.

However, the projectile goes up, reaches a maximum height, and then comes down. It hits 80 feet once on the way up (earlier time) and once on the way down (later time).

Since $\sqrt{5} \approx 2.24$:

• $\dfrac{3 + \sqrt{5}}{2} \approx \dfrac{5.24}{2} \approx 2.62$ seconds (later time),
• $\dfrac{3 - \sqrt{5}}{2} \approx \dfrac{0.76}{2} \approx 0.38$ seconds (earlier time).

We are asked for when it first reaches 80 feet, so we choose the smaller positive time:

This matches answer choice D.