Hey there!
I solved this problem using the quadratic formula however I'm sure that you can use other methods of solving a quadratic.
As we are trying to figure out the time that the ball hits the ground we can set \(h\) as equal to \(0\).
\(0=-16^2-24t+180 \)
I decided to divide all of our terms by -4 to give us some easier terms to work with.
\(0=4t^2+6t-45\)
Plugging our numbers into the quadratic formula of \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
\(\frac{-6\pm\sqrt{6^2-(4)(4)(-45)}}{2(4)}\)
After evaluating we get the answer of \(\frac{-3\pm3\sqrt{21}}{4}\)
This can be put in decimal form (rounded to the nearest hundred) as 2.69, or -4.19
Since your answer can't be negative, the ball will hit the ground after 2.69 seconds.
Hope this helps!