Seven iterations of Newton's method, which in this case gave;

(3x^4+4x^3-x^2+3)/(4x^3+6x^2-2x-6) ≈ 1.6180339887499... and -0.6180339887499

using 1 and -1 as starting values of x.

My TI-89 gave (√(5) + 1)/2 and -(√(5) - 1)/2 ( which agrees with above )

Solving your equation for zero gives;

x^4 + 2x^3 - x^2 - 6x - 3 = 0

I don't know the algorithm used by the TI-89 so I had to use Newton's method. I would think that a quartic formula would be more cumbersome than Newton's method, but I am sure that you can find it on the internet easily. In short, I don't see an easy way to solve this.

]]>And that's a quartic equation. They can be solved, but the method is very complex. I think I've probably taken the wrong route. That or the quartic can be factorised into something nicer.

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