so, h(t) is how far is Daniel's head from the surface of the water, namely the surface itself is when the height is nill, so his head is at the surface and the height of it is just 0, thus h(t) = 0. Namely, what is "t" when h(t) is 0?
[tex]\bf h(t)=-8t^2-28t+60\implies \stackrel{h(t)}{0}=-8t^2-28t+60
\\\\\\
0=-2t^2-7t+15\implies 2t^2+7t-15=0\implies (2t+3)(t-5)=0
\\\\\\
\begin{cases}
2t+3=0\implies 2t=-3\implies &t=-\frac{3}{2}\\\\
t-5=0\implies &\boxed{t=5}
\end{cases}[/tex]
clearly the seconds cannot be a negative unit, so is not -3/2.
