4. Solve
[tex]1430 = 0.55t^2 + 550[/tex]
[tex] 1430 - 550 =880 = 0.55t^2[/tex]
[tex]880/0.55 = 1600 = t^2[/tex]
[tex]t= \pm \sqrt{1600} = \pm 40[/tex]
On the positive answer makes sense.
Answer: Month 40
5. The throw ends when the height is zero (I guess we're ignoring rolling).
[tex]0 = -0.02x^2 + x + 6[/tex]
[tex] x = \dfrac{-1 \pm \sqrt{1^2 - 4(6)(-0.02)}}{2(-0.02)}[/tex]
Only the negative sign for the plus/minus will give a sensible, positive answer:
[tex]x = \dfrac{-1 - \sqrt{1^2 - 4(6)(-0.02)}}{2(-0.02)} = 55.4138[/tex]
Answer: 55.4 feet
