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Question 423230: (20t^2+t)^2-22(20t^2+t)+21=0
what are the solutions of t?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! (20t^2+t)^2-22(20t^2+t)+21=0
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Let w = 20t^2+t
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Substitute to get:
w^2 - 22w + 21 = 0
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Factor:
(w-21)(w-1) = 0
w = 21 or w = 1
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Solve for "t":
If w = 21,
20t^2+t = 21
20t^2+t-21 = 0
(t-1)(20t+21) = 0
t = 1 or t = -21/20
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If w = 1
20t^2+t= 1
20t^2+t-1 = 0
(5t-1)(4t+1) = 0
t = 1/5 of t = -1/4
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Cheers,
Stan H.
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