The given expression is
[tex]h=32-2t-5t^2[/tex]The ground is at h = 0, so
[tex]0=32-2t-5t^2[/tex]Now, we use the quadratic formula to find the solutions.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where a = -5, b = -2, and c = 32.
[tex]\begin{gathered} x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4\cdot(-5)\cdot(32)}}{2\cdot(-5)}=\frac{2\pm\sqrt[]{4+640}}{-10}=\frac{2\pm\sqrt[]{644}}{-10} \\ x=\frac{2\pm25.38}{-10} \\ x_1=\frac{2+25.38}{-10}=-2.7 \\ x_2=\frac{2-25.38}{-10}=2.34 \end{gathered}[/tex]In this case, just the positive solution makes sense.
