SOLUTION: Solve the equation: (t-4)^6-11(t-4)^3+28=0 t=

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Question 1033563: Solve the equation:
(t-4)^6-11(t-4)^3+28=0
t=

Found 2 solutions by stanbon, josgarithmetic:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the equation:
(t-4)^6-11(t-4)^3+28=0
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This is a quadratic equation with variable (t-4)^3
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Let W = (t-4)^3 ; then you get:
W^2 - 11W + 28 = 0
Factor::
(W-7)(W-4) = 0
W = 7 or W = 4
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Solve for "t"::
(t-4)^3 = 7 or (t-4)^3 = 4
t = 7^(1/3) + 4 or t = 4^(1/3) + 4
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Cheers,
Stan H.
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Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Do you see this is in quadratic form?

Let v=t-4------------even better, let v=%28t-4%29%5E3. You would use this to first solve for v.

v%5E2-11v%2B28=0