Landon is standing in a hole that is 5.1 ft deep. He throws a rock, and it goes up into the air, out of the hole, and then lands on the ground above. The path of the rock can be modeled by the equation y = -0.005x2 + 0.41x - 5.1, where x is the horizontal distance of the rock, in feet, from Landon and y is the height, in feet, of the rock above the ground. How far horizontally from Landon will the rock land?

Refer to the diagram shown below.

The path of the rock is 
y = -0.005x² + 0.4x - 5.1

When x = 0, y = -5.1 ft, which is the location of the hole level.
The rock reaches ground level two times when x = x₁ and when x = x₂.

At ground level, y = 0. Therefore
-0.005x² + 0.4x - 5.1 = 0
Divide through by  -0.005.
x² - 82x + 1020 = 0

Solve with the quadratic formula.
x = (82 +/- √(82² - 4*1020))/2
   = (82 +/- 51.4198)/2
x = 66.71 or x = 15.29

The smaller value of x is x₁ = 15.3 ft when the rock emerges out of the hole and reaches ground level.
The larger value of x is x₂ = 66.7 ft when the rock falls to the ground.
The rock lands 66.7 ft horizontally from Landon.

Answer:  66.7 ft