Question #7b374

1 Answer
Nam D.
Jan 8, 2018

#x~~0.86887...#

Explanation:

There is not really an algebraic way to do this, besides Newton's method.

I used Symbolab (an online math engine like Wolfram-Alpha), and got #x~~0.86887...#.

An alternative way would be to graph #-6x^3-18x^2-4x+21=0#. graph{-6x^3-18x^2-4x+21 [-5.59, 4.41, -3.98, 1.02]}

If click on around #x=0.869#, it comes up with #y=0#, so #x=0.869# would also be a value.

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