Chapter 5 - Polynomials and Factoring - 5.8 Solving Applications - 5.8 Exercise Set - Page 362: 45

1,2

We are given the following equation: $$ h=48t-16t^2$$ Thus, plugging in $32$ for $h$, we find: $$32=48t-16t^2 \\ -16t^2+48t-32=0 \\ t_{1,\:2}=\frac{-48\pm \sqrt{48^2-4\left(-16\right)\left(-32\right)}}{2\left(-16\right)} \\ t=1,2$$