Answer
Algebraically, the year is $2010$.
Graphically, the year is $2010$.
The graph is as shown.
Work Step by Step
Verifying algebraically:
$$D=-0.071t^2+2.94t-10.0$$
Substituting $D=13$ into the equation:
$$13=-0.071t^2+2.94t-10.0$$
Solving for $t$:
$$-71t^2+2940-10000=13000$$
$$-71t^2+2940t-23000=0$$
Using the quadratic formula:
$$t=\frac{-2940\pm\sqrt{2940^2-4(-71)(-23000)}}{2(-71)}=\frac{1470\pm10\sqrt{5279}}{71}$$
$$t_1=\frac{1470-10\sqrt{5279}}{71}=10.47$$
$$t_2=\frac{1470+10\sqrt{5279}}{71}=30.94$$
With the constraint of $8\leq t\leq14$, take $t=10.47$. As $t=10.47$ belongs in year $10$, take $t=10$.
Then, the year is:
$$Year=2000+10=2010$$
Verifying graphically, as shown, notice that at $D=13$, the $t=10.47$.
Thus, similarly, the year is $2000+10=2010$.
