Question

A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation

Answer 1

Ok

y=-0.04x^2+8.3x+4.3
when the rocket reaches the ground (when height=0, ie when y=0), then the rocket will land, find the x coordinate

set y=0
0=-0.04x^2+8.3x+4.3
use quadratic formula
if you have ax^2+bx+c=0, then
x=$\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$
a=-0.04
b=8.3
c=4.3
x=$\frac{-8.3+/- \sqrt{8.3^{2}-4(-0.04)(8.3)} }{2(-0.04)}$
x=208.017 or -0.516785
xrepresents horizontal distance
you cannot have a negative horizontal distance unless you fired and theh wind blew it backwards
therefor x=280.017 is the answer
208.02 m