a model rocket is launched from roof into a large field the path of the rocket can be modeled by the equation y= -0.02x^2+2.3x+6 where x is the horizontal distance ,in meters,from the starting point on the roof and y is the height in meters,of the rocket above the ground . how far horizontally form its starting point will the rocket land?round your answer to the nearest hundredth

57.50m
115.00m
117.55m
235.10m

117.55? Thanks Reiny!!

Its C 117.55m

all anwers ??????????

when the rocket lands ,

-0.02x^2+2.3x+6 = 0
0.02x^2 - 2.3x - 6 = 0
multiply by 100
2x^2 - 230x - 600 = 0 , "nicer" numbers
use the quadratic formula, using the positive answer
I see it in your choices.

To find the horizontal distance from the starting point where the rocket lands, we need to find the value of x when y is equal to zero. This is because when the rocket lands, its height above the ground would be zero.

We can set the equation equal to zero and solve for x:
0 = -0.02x^2 + 2.3x + 6

To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -0.02, b = 2.3, and c = 6. Plugging these values into the quadratic formula, we get:
x = (-2.3 ± √(2.3^2 - 4(-0.02)(6))) / (2(-0.02))

Simplifying the expression further:
x = (-2.3 ± √(5.29 + 0.48)) / (-0.04)
x = (-2.3 ± √5.77) / (-0.04)

Now we can calculate both possible values of x:
x1 = (-2.3 + √5.77) / (-0.04)
x2 = (-2.3 - √5.77) / (-0.04)

Calculating x1 and x2 using a calculator:
x1 ≈ -57.50
x2 ≈ 117.55

Since we're looking for the positive value of x, the rocket will land approximately 117.55 meters horizontally from its starting point. Therefore, the correct answer is 117.55m.