solve for t and solve using quadratic formula h(t) = -16t^2 + 150t + 50
To solve the equation h(t) = -16t^2 + 150t + 50, we need to set it equal to zero and find the values of t.
-16t^2 + 150t + 50 = 0
Now, we can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where a, b, and c are the coefficients in the equation -16t^2 + 150t + 50 = 0.
In this case:
a = -16
b = 150
c = 50
Substituting the values into the quadratic formula:
t = (-150 ± √(150^2 - 4(-16)(50))) / (2(-16))
Simplifying further:
t = (-150 ± √(22500 + 3200)) / (-32)
t = (-150 ± √(25700)) / (-32)
t = (-150 ± 160.311) / (-32)
Now, we can calculate the two solutions:
t1 = (-150 + 160.311) / (-32)
= 10.31 / (-32)
= -0.322
t2 = (-150 - 160.311) / (-32)
= -310.311 / (-32)
= 9.696
Thus, the two solutions for t are approximately t = -0.322 and t = 9.696.
is this right?
and are you sure?
Apologies for the confusion. Let's go through the calculations again to ensure accuracy.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In the given equation:
a = -16, b = 150, c = 50
t = (-150 ± √(150^2 - 4(-16)(50))) / (2(-16))
Simplifying further:
t = (-150 ± √(22500 + 3200)) / (-32)
t = (-150 ± √(25700)) / (-32)
t = (-150 ± 160.311) / (-32)
Now, calculating the solutions:
t1 = (-150 + 160.311) / (-32)
= 10.311 / (-32)
= -0.322
t2 = (-150 - 160.311) / (-32)
= -310.311 / (-32)
≈ 9.696
Therefore, the corrected solutions for t are approximately t = -0.322 and t ≈ 9.696.