The height h in feet of a baseball above the ground t second

The height h, in feet of a baseball above the ground t seconds after it is hit, is given by h= -16t^2 + 50t +4.5
Use this equation to determine the number of seconds, to the nearest tenth of a second, from the time the ball is hit until the ball hits the ground.
Show all of your algebra work.

Solution

h = -16t^2 + 50t + 4.5
when the ball hits the ground, h will be 0
so solve the equation
-16t^2 + 50t + 4.5 = 0
solve using quadratic equation (-b +/- sqrt(b^2-4ac))/2a.
a = -16
b = 50
c = 4.5
use calculator and do not round until the end.
b^2 = 2500
4ac = 4*(-16)*4.5 = (-288)
b^2-4ac = 2500 - (-288) = 2788
sqrt(b^2-4ac) = 52.80151513
formula then becomes (-50 +/- 52.80151513) / -32
t = 3.212547345 or t = .087547348
substituting in original equation for t = 3.212547345
(-16 * 10.32046046) + (50 * 3.212547345) + 4.5 = something very close to 0.
substituting in original equation for t = .087547345
(-16 * .007) + (50 * .087) + 4.5 = something very positive so this is not the answer.
answer is t = 3.2 seconds rounded to the nearest tenth of a second.