check the picture below, so it hits the ground when h = 0, therefore,
[tex]\bf \stackrel{h}{0}=160-12t-16t^2\implies 16t^2+12t-160=0\\\\\\ 4t^2+3t-40=0
\\\\\\
\textit{now let's use the quadratic formula}
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
y=& 4 t^2& +3 t& -40\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad \qquad
t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}[/tex]
[tex]\bf t=\cfrac{-3\pm\sqrt{3^2-4(4)(-40)}}{2(4)}\implies t=\cfrac{-3\pm\sqrt{9+640}}{8}
\\\\\\
t=
\begin{cases}
\frac{-3+\sqrt{649}}{8}\\\\
\frac{-3-\sqrt{649}}{8}
\end{cases}\approx
\begin{cases}
\boxed{2.81}\\\\
-3.56
\end{cases}[/tex]
