[tex]\bf h=-9.8t^2+45t+1.2\implies 64.5=-9.8t^2+45t+1.2
\\\\\\
0=-9.8t^2+45t-63.3\\\\
-------------------------------[/tex]
[tex]\bf \qquad \qquad \qquad \textit{discriminant of a quadratic}
\\\\\\
0=\stackrel{\stackrel{a}{\downarrow }}{-9.8}t^2\stackrel{\stackrel{b}{\downarrow }}{+45}t\stackrel{\stackrel{c}{\downarrow }}{-63.3}
~~~~~~
\stackrel{discriminant}{b^2-4ac}=
\begin{cases}
0&\textit{one solution}\\
positive&\textit{two solutions}\\
negative&\textit{no solution}
\end{cases}
\\\\\\
(45)^2-4(-9.8)(-63.3)\implies -456.36[/tex]
the discriminant is a negative value, thus no solution for such quadratic, meaning if we use h=64.5 like we did, there's no "t" seconds at which point the ball hits the batting cage.
