1. The problem is to analyze the quadratic function $h(t) = -5t^2 + 20t + 2$. 2. The general form of a quadratic function is $ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 3. Here, $a = -5$, $b = 20$, and $c = 2$. 4. To find the vertex (maximum or minimum), use the formula for the $t$-coordinate of the vertex: $$t = -\frac{b}{2a} = -\frac{20}{2 \times -5} = 2.$$ 5. Substitute $t=2$ back into $h(t)$ to find the vertex value: $$h(2) = -5(2)^2 + 20(2) + 2 = -20 + 40 + 2 = 22.$$ 6. Since $a = -5 < 0$, the parabola opens downward, so the vertex at $(2, 22)$ is a maximum point. 7. To find the roots (where $h(t) = 0$), use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{400 - 4(-5)(2)}}{2(-5)} = \frac{-20 \pm \sqrt{400 + 40}}{-10} = \frac{-20 \pm \sqrt{440}}{-10}.$$ 8. Simplify the square root: $$\sqrt{440} = \sqrt{4 \times 110} = 2\sqrt{110}.$$ 9. So the roots are: $$t = \frac{-20 \pm 2\sqrt{110}}{-10} = 2 \mp \frac{\sqrt{110}}{5}.$$ 10. These are the two points where the function crosses the $t$-axis. Final answer: The vertex is at $(2, 22)$ (maximum), and the roots are $t = 2 \pm \frac{\sqrt{110}}{5}$.