Question 1130410
When a stopper is removed from the bottom of a barrel filled with water, the depth d, in centimeters, of a liquid in the barrel can be approximated by d=0.039t^2-5.816t+200, where t is the time since stopper was removed from the hole. When will the depth be 125 cm? Round to the nearest tenth of a second. 
So far I have:
125=0.039t^2-5.816t+200 then I subtracted 125 from both sides so the equation is equal to 0 which will make it 0=0.039t^2-5.816t+75 I know I need to factor it but I'm not sure how to go about that. 
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0.039t^2-5.816t+75 = 0
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It might be factorable, might not.
Use the quadratic equation:
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*[invoke solve_quadratic_equation 0.039,-5.816,75]
It cannot be factored, but there's another problem.
It's a parabola that opens upward, giving 2 solutions for t.
Check your equation for d.