I need help with this math problem?

H(t) = -2^2+19t+6 = 0

solve for t

It looks like "the top of a school" is 6 units above ground level, since the starting height H(0) = 6. Assuming it's going to land on the ground, you need to find the time t>0 when H(t) = 0. If that "t2" means "t squared", the conventional way to type that is "t^2" with ^ indicating that an exponent follows.

-2t^2 + 19t + 6 = 0

Use the quadratic formula:

t = { -19 +/- sqrt[ 19^2 - 4(-2)(6)) ] } / [2 * (-2)]

t = 19/4 +/- sqrt(19^2 + 48) / 4

Subtracting will give you a negative value for t, so that's out. Adding gives a value of about 9.8 for t.

Essentially, you want to solve for h = 0

so

0 = -2t^2 + 19t + 6

==> t = .....

Use the quadratic formula: for ax^2 + bx + x = 0

x = (-b +/- sqrt(b^2 - 4ac))/2a

so

t = (-19 -sqrt(19^2 + 4*2*6))/(-4) (the other solution is less than 0, so ignore it)

which is about 39 /4 = 9.75 seconds ==> ans = 10

Oh cool! I just learned about this in calculus. That quadratic is the position graph, and to know the flight time, you’ll need the velocity graph. So take the derivative of the quadratic. That will model the rocket’s velocity. Next solve the velocity equation for zero, since the only time the rockets velocity is zero is at the top of its arc. Then you multiply the flight time by 2 and account for the if the rocket landed at a different height than it started. Hope this helps!