Question 2 of 15, Step 1 of 1 Correct Pleasantburg has a population growth model of $P(t)=a t^{2}+b t+P_{0}$ where $P_{0}$ is the initial population. Suppose that the future population of Pleasantburg $t$ years after January 1,2012 , is described by the quadratic model $P(t)=0.9 t^{2}+6 t+26,000$. In what month and year will the population reach 39,000 ? transcript Answer Keypad Keyboard Shortcuts

Step 1 ：\(39,000 = 0.9t^2 + 6t + 26,000\)

Step 2 ：\(13,000 = 0.9t^2 + 6t\)

Step 3 ：\(14,444.44 = t^2 + 6.67t\)

Step 4 ：\(t^2 + 6.67t - 14,444.44 = 0\)

Step 5 ：\(t = \frac{-6.67 \pm \sqrt{(6.67)^2 - 4*1*(-14,444.44)}}{2*1}\)

Step 6 ：\(t = \frac{-6.67 \pm \sqrt{44.49 + 57,777.76}}{2}\)

Step 7 ：\(t = \frac{-6.67 \pm \sqrt{57,822.25}}{2}\)

Step 8 ：\(t = \frac{-6.67 \pm 240.46}{2}\)

Step 9 ：\(t = \frac{240.46 - 6.67}{2} = 116.89 \text{ years}\)

Step 10 ：\(t = \frac{-240.46 - 6.67}{2} = -123.56 \text{ years}\)

Step 11 ：\(\text{Discard the negative solution, so } t = 116.89 \text{ years}\)

Step 12 ：\(\text{Add } 116.89 \text{ years to the start year 2012, so the year is } 2012 + 116 = 2128\)

Step 13 ：\(\text{Calculate the month: } 0.89*12 = 10.68, \text{ round up to 11}\)

Step 14 ：\(\boxed{\text{Therefore, the population of Pleasantburg will reach 39,000 in November, 2128}}\)