a ball is thrown upward with a velocity of 50 feet per second from the top edge of a building 20 feet high. for how long is the ball higher

 This could also be solved by putting p(t) = 50 and solving for t.  There will be two values for t (one for the ball on its way up and one for it on its way down).  The difference between them gives the time you are looking for.

The equation can be written as : 50 = -16t² + 50t + 20 or:

16t² - 50t + 30 = 0 

$${\mathtt{16}}{\mathtt{\,\times\,}}{{\mathtt{t}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{50}}{\mathtt{\,\times\,}}{\mathtt{t}}{\mathtt{\,\small\textbf+\,}}{\mathtt{30}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{145}}}}{\mathtt{\,-\,}}{\mathtt{25}}\right)}{{\mathtt{16}}}}\\
{\mathtt{t}} = {\frac{\left({\sqrt{{\mathtt{145}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{25}}\right)}{{\mathtt{16}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{t}} = {\mathtt{0.809\: \!900\: \!338\: \!825\: \!481\: \!5}}\\
{\mathtt{t}} = {\mathtt{2.315\: \!099\: \!661\: \!174\: \!518\: \!5}}\\
\end{array} \right\}$$

Giving the same result as found graphically by Chris

.

Have a look at the graph of the function, here.....

Graph

Notice that the ball is higher than 50 ft between about .81 seconds and 2.315 seconds. So,  (2.315 - .81) = 1.505 secs