Question 1109935
AN AIRPLANE FLYING EAST AT THE RATE OF 6 MILES/MINUTES ,PASSED OVER A COURT-HOUSE AT 2 PM. A SECOND PLANE FLYING NORTH AT THE RATE OF 6 MILES/MINUTES,PASSED OVER THE COURT-HOUSE AT 2:04 PM. IV THE PLANES ARE FLYING AT THE SAME  ALTITUDE,IN HOW MANY MINUTES AFTER 2 PM WILL THEY BE 36 MILES APART?
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p = distance of the 1st plane from the courthouse at t minutes past 2
q = distance of the 2nd plane from the courthouse at t minutes past 2
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t=0 at 2:04
p = 6t + 4*6 = 6t+24
q = 6t
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(6t+24)^2 + (6t)^2 = 36^2
(t+4)^2 + (t)^2 = 36
2t^2 + 8t -20 = 0
*[invoke solve_quadratic_equation 2,8,-20]
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t =~ 1.74 minutes past 2:04
--> 5.74 minutes past 2:00