Answer
a. Jan 1, 2019.
b. $t=19^{th}$ year
Work Step by Step
$F=1000(30+17t-t^2)$,
a. $t=0$ for Jan 1, 2002.
$t=0, F=1000(30)=30,000$.
Thus, $F=1000(30+17t-t^2)=30,000$,
$17000t-t^2=0$,
$17t-t^2=0$,
$t(17-t)=0$,
$t=0$ or $t=17$.
Therefore, The answer is for Jan 1, 2019.
b.$F=1000(30+17t-t^2)=0$,
$30000+17000t-1000t^2=0$,
$30+17t-t^2=0$ multiplying both sides by -1.
$t^2-17t-30=0$
Solving for the trinomial using quadratic formula for the quadratic equation of $ax^2+bx+c, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $t^2-17t-30=0, x=\frac{17\pm\sqrt{(-17)^2-4(1)(-30)}}{2(1)}=\frac{17\pm20.2}{2}$
thus, $t=18.6$ or $t=-1.6$. Since we're solving for time, only positive answers are relevant.
Therefore, rounding to the nearest year, $t=19^{th}$ year
