SOLUTION: Solve. (2t^2+t)^2-4(2t^2+t)+3=0. Let n=(2t^2+t).

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Question 937325: Solve. (2t^2+t)^2-4(2t^2+t)+3=0. Let n=(2t^2+t).
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve. (2t^2+t)^2-4(2t^2+t)+3=0. Let n=(2t^2+t).
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n^2 - 4n + 3 = 0
Factor::
(n-3)(n-1) = 0
n = 3 or n = 1
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Solve for "t"::
2t^2+t = 3 or 2t^2+t = 1
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2t^2+t-3 = 0 or 2t^2+t-1 = 0
(2t+3)(t-1) = 0 or t = [-1+-sqrt(1-4*2*-1)]/4
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t = -3/2 or t = 1 or t = (-1+3)/4 or t = -4/4
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t = -3/2 or t = 1 or t = 1/2 or t = -1
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Cheers,
Stan H.
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