If the stone strikes the ground 65 m away, find the time of flight and initial speed?

Vxi = 17.7 m/s

Vyi = 8.636 m/s

xi = yi = 0, yf = -65.0 m, ay = -g, and vi = 19.7 m/s

This is my work so far:
trying to find t: 4.9t^2 - 8.636 -65=0

Then I used the quadratic formula and got t = 4.6285 but it's not right.

To find the time of flight and initial speed of the stone, we can use the equations of motion in the vertical direction.

First, let's find the time of flight using the equation:
yf = yi + Vyi * t + (1/2) * a * t^2

Substituting the given values:
-65.0 = 0 + 8.636 * t + (1/2) * (-9.8) * t^2
Rearranging the equation:
0 = -4.9t^2 + 8.636t - 65.0

To solve this quadratic equation, you can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)

Here, a = -4.9, b = 8.636, and c = -65.0.
Plugging in these values:
t = (-8.636 ± sqrt((8.636^2) - 4*(-4.9)(-65.0))) / (2*(-4.9))

Evaluating the formula, you will get two possible values for t. Take the positive value since time cannot be negative.

Now, let's find the initial speed (vi) of the stone using the equation:
Vx = Vxi

Substituting the given value:
Vxi = 17.7 m/s

Therefore, the time of flight (t) and initial speed (vi) can be calculated by solving the quadratic equation and using the given value, respectively.