For (b), consider the discriminant of the equation

(Remember that you need to rearrange to get $0$ on one side before you can use the discriminant.)

For (c), the value of the car in the year $2100$ is given when $t=80$.

1. When $t=0,V=\pounds 20000$

2. Suppose that $V=4500$. Then

   $$\begin{array}{rl}7t^2-650t+20000& =4500\\ 7t^2-650t+15500& =0\end{array}$$

   Now consider the discriminant

   $$\begin{array}{rl}\mathrm{\Delta }& ={650}^2-4(7)(-15500)\\ & =-11500\\ & <0\end{array}$$

   Since the discriminant is negative, there is no real solution, and so no time $t$ at which the value of the car is $\pounds 4500$.

3. The year $2010$ occurs when $t=80$, so the value is

   $$\begin{array}{rl}V& =7(8{)}^2-650(8)+20000\\ & =\pounds 12800\end{array}$$

   Thus, we need to solve

   $$\begin{array}{rl}7t^2-650t+20000& =12800\\ 7t^2-650t+7200& =0t& =\frac{650\pm \sqrt{{650}^2-4(7)(7200)}}{2(7)}\\ & =\frac{650\pm 220900}{14}\end{array}$$

   We are unsurprised to see that one of the solutions is $t=80$. We are interested in the other solution, $t=12.857\dots$.

   So the car will attain the value $\pounds 12800$ at some point in the year $2032$.