Not getting the correct answer

Question

y = 0.343t2+ 11.74t+5.17 where t = Time of day (hour) and y = Temperature (in degrees)
The times are the hours since midnight. For instance, 7 means 7 am, and 13 means 1 pm.
(a)Using algebraic techniques we have learned, find the maximum temperature predicted by the quadratic model and find the time when it occurred. Report the time to the nearest quarter hour (i.e., __:00 or __:15 or __:30 or __:45). (For instance, a time of 18.25 hours is reported as 6:15 pm.) Report the maximum temperature to the nearest tenth of a degree.
(b) Use the quadratic polynomial to estimate the outdoor temperature at 7:30 am, to the nearest tenth of a degree. (work optional)
(c) Use the quadratic polynomial y = 0.343t2+ 11.74t+5.17 together with algebra to estimate the time(s) of day when the outdoor temperature y was 100 degrees. That is, solve the quadratic equation 100 = 0.343t2+ 11.74t+5.17 Show algebraic work in solving. State your results clearly; report the time(s) to the nearest quarter hour.

Answer 1

(a). Since we are to maximize temperature, I assume that the small square to the left of 0.343 is supposed to be a                minus sign?

 

      If so, then y = -0.343t² + 11.74t + 5.17

 

      Maximum temp occurs when t = -11.74 / [2(-0.343)] = 17.113703 hours after midnight

 

                                                                                    ≈ 5:00 PM

 

       Maximum temperature = 105.6°

 

(b).  At 7:30 AM, t = 7.5

 

        Temp = 73.9°

 

(c).  -0.343t² + 11.74t + 5.17 = 100

 

       -0.343t² + 11.74t - 94.83 = 0

 

        t = [-11.74 ± √7.72084] / (-0.686) = 13.0632 or 21.164

 

        t = 1:00 PM or 9:00 PM