Check the picture below.
[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+80}t\stackrel{\stackrel{c}{\downarrow }}{+96} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
[tex]\left(-\cfrac{ 80}{2(-16)}~~~~ ,~~~~ 96-\cfrac{ (80)^2}{4(-16)}\right) \implies \left( - \cfrac{ 80 }{ -32 }~~,~~96 - \cfrac{ 6400 }{ -64 } \right) \\\\\\ \left(\cfrac{5}{2}~~,~~96+100 \right)\implies \stackrel{highest~point}{\stackrel{\qquad ~~ \downarrow }{\left(2\frac{1}{2}~~,~~196 \right)}}[/tex]
