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From: Axel Vogt <&noreply@axelvogt.de>
Newsgroups: comp.soft-sys.math.maple
Subject: Re: Finding all Solutions
Date: Tue, 06 Jan 2015 21:59:23 +0100
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On 06.01.2015 11:26, Thomas D. Dean wrote:
> I have two equations,
>
> eq1:=y=-2*x^2 + 10*x - 6;
> eq2:=y=6*x^3 + 5*x^2 - 8;

Besides what was said/written: this is for same y,
so -2*x^2 + 10*x - 6 =  6*x^3 + 5*x^2 - 8. Bring it
to 'one side' to have a cubic = p.

Plot it in your expected range. In x=0 and x=-1 one
can evaluate and as p goes to +-oo for x --> +-oo
it is easy to see that there must be 2 real zeros.

It follows (from algebra) that the 3rd zero is real
as well. Which justifies the usage of Real methods.