problem: A ball travels on a parabolic path in which the height (in feet) is given by the expression $-25t^2+75t+24$, where $t$ is the time after launch. At what time is the height of the ball at its maximum?
solution: First, we find the maximum height of the ball by maximizing the expression $-25t^2+75t+24$. We will do this by completing the square. Factoring a $-25$ from the first two terms, we have  \[-25t^2+75t+24=-25(t^2-3t)+24\]To complete the square, we add and subtract $\left( -\frac{3}{2}\right)^2=\frac{9}{4}$ inside the parentheses to get \begin{align*}
  -25(t^2-3t)+24&=-25\left(t^2-3t+\frac{9}{4}-\frac{9}{4}\right)+24\\
  &=-25\left(\left(t-\frac{3}{2}\right)^2-\frac{9}{4}\right)+24\\
  &=-25\left(t-\frac{3}{2}\right)^2+\frac{225}{4}+\frac{96}{4}\\
  &=-25\left(t-\frac{3}{2}\right)^2+\frac{321}{4}
  \end{align*}Since $-25\left(t-\frac{3}{2}\right)^2$ is always non-positive, the maximum value of the expression is achieved when $-25\left(t-\frac{3}{2}\right)^2=0$. This happens when $t-\frac{3}{2}=0$. Therefore the height of the ball is at its maximum when $t=\boxed{\frac{3}{2}}$.