Chapter 9 - Solving Quadratic Equations - 9.5 - Solving Quadratic Equations Using the Quadratic Formula - Exercises - Page 521: 23

$t\approx0.22~s~\text{and}~t\approx1.40~s$

$$h=-16t^2+26t$$ Substitute $h=5$: $$5=-16t^2+26t$$ $$-16t^2+26t-5=0$$ $$t=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$$ $$t=\frac{-26\pm\sqrt {(26)^2-4(-16)(-5)}}{2(-16)}=\frac{-26\pm\sqrt {356}}{-32}\approx\frac{-26\pm18.87}{-32}$$ $$t\approx\frac{-26+18.87}{-32}\approx0.22~~~~\text{and}~~~~t\approx\frac{-26-18.87}{-32}\approx1.40$$