The equation y = 16t^2  60t + 54$ describes the height (in feet) of a ball thrown downward at 60 feet per second from a height of 54 feet from the ground. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.
height = 16t^{2}  60t + 54 , where t is the number of seconds after the ball is released
In how many seconds will the ball hit the ground?
In other words, what is t when the height is zero?
Plug in 0 for the height and solve for t .
0 = 16t^{2}  60t + 54
We can divide both sides of the equation by 2
0 = 8t^{2} + 30t  27
Let's split 30t into two terms such that their coefficients multiply to 216
0 = 8t^{2}  6t + 36t  27
Factor 2t out of the first two terms and factor 9 out of the last two terms
0 = 2t(4t  3) + 9(4t  3)
Factor (4t  3) out of both remaining terms
0 = (4t  3)(2t + 9)
Set each factor equal to 0 and solve for t
4t  3 = 0  _______or_______  2t + 9 = 0 

4t = 3 
 2t = 9  
t = 3/4  t = 9/2 
 
t = 0.75  t = 4.5 
We want the number of seconds after the ball is released, so we only want the positive solution.
The ball will hit the ground 0.75 seconds.
Here's a graph: https://www.desmos.com/calculator/kznk4jayc4
Yep! You can use the quadratic formula to solve 16t^{2}  60t + 54 = 0 for t , like this:
\(t \,=\, {(60) \,\pm \,\sqrt{(60)^24(16)(54)} \over 2(16)}\,=\,{60\, \pm \,\sqrt{7056} \over 32}\,=\, {60 \,\pm\, 84 \over 32}\\~\\ \ \\ \begin{array}\ t\,=\,\frac{60+84}{32}&\qquad\text{or}\qquad& t\,=\,\frac{6084}{32}\\~\\ t\,=\,\frac{144}{32}&\qquad\text{or}\qquad& t\,=\,\frac{24}{32}\\~\\ t\,=\,4.5&\qquad\text{or}\qquad& t\,=\,0.75 \end{array}\)