1. **State the problem:** We are given the quadratic function $h(t) = -5t^2 + 20t + 2$ and need to analyze it. 2. **Formula and rules:** This is a quadratic function of the form $h(t) = at^2 + bt + c$ where $a = -5$, $b = 20$, and $c = 2$. The graph is a parabola. 3. **Find the vertex:** The vertex $t$-coordinate is given by $t = -\frac{b}{2a} = -\frac{20}{2(-5)} = 2$. 4. **Calculate the vertex height:** Substitute $t=2$ into $h(t)$: $$h(2) = -5(2)^2 + 20(2) + 2 = -5(4) + 40 + 2 = -20 + 40 + 2 = 22$$ 5. **Interpretation:** The vertex is at $(2, 22)$, which is the maximum point since $a < 0$. 6. **Find the roots:** Solve $-5t^2 + 20t + 2 = 0$ using the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{20^2 - 4(-5)(2)}}{2(-5)} = \frac{-20 \pm \sqrt{400 + 40}}{-10} = \frac{-20 \pm \sqrt{440}}{-10}$$ 7. **Simplify roots:** $$\sqrt{440} = \sqrt{4 \times 110} = 2\sqrt{110}$$ So, $$t = \frac{-20 \pm 2\sqrt{110}}{-10} = \frac{-20}{-10} \pm \frac{2\sqrt{110}}{-10} = 2 \mp \frac{\sqrt{110}}{5}$$ 8. **Final roots:** $$t_1 = 2 + \frac{\sqrt{110}}{5}, \quad t_2 = 2 - \frac{\sqrt{110}}{5}$$ **Summary:** The parabola opens downward with vertex at $(2, 22)$ and roots at $t = 2 \pm \frac{\sqrt{110}}{5}$.