B) Given
[tex]y=0.2313x^2+2.600x+35.17[/tex]
Set y=65 and solve for x, as shown below
[tex]\begin{gathered} y=65 \\ \Rightarrow0.2313x^2+2.600x+35.17=65 \\ \Rightarrow0.2313x^2+2.600x-29.83=0 \end{gathered}[/tex]
Then, solve the quadratic equation using the quadratic formula,
[tex]\begin{gathered} x=\frac{-2.600\pm\sqrt{(2.600)^2-4*0.2313*-29.83}}{2*0.2313} \\ \Rightarrow x=-18.2915,7.05065 \end{gathered}[/tex]
The function is not valid for negative values of x; therefore, the solution can only be x=7.05065 which can be rounded to x=7. Furthermore, x=7 corresponds to the year 2001. The answer to part B is 2001.