Alan

x = pi is also a solution

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Plotting a graph often helps:

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3 is a solution

$${{\mathtt{3}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{10}}$$

(It's the only real solution)

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You can use the Solve equations function to find the coefficients of a polynomial that fits through a number of points and then plot the resulting function in MM4.

For example, suppose you wanted to fit y = a.x² + b.x + c through the points (-2, 3), (1, 6) and (2, 15)

You can do this with MM4 (though, annoyingly, you have to use x1, x2 and x3 for the unknown coefficients, whereas you really would like to use a, b and c so as not to confuse the unknowns with values on the x-axis).

Having done this you can plot the graph of the resulting function (which you have to construct manually, now you know the coefficients):

I didn't bother turning off the axes in the above, although that can be done of course.

Well, the Microsoft one can do your first condition (see below), though I'm not sure about your second one (do you mean you want the calculator to find a curve that passes through the specified points and then plot the curve?).

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The 64 one will only work on 64-bit computers.  The other one will work on both 32 and 64-bit computers. 

I've only recently come across this calculator, so I've been trying it out.  It's ok, though it does have a few annoying features (for example, changing the colour of lines on a graph is cumbersome), and it isn't powerful enough for general "industrial" use, but it should be very useful in an educational environment.  One useful feature, for example, is that after you solve some equations you have the option to examine the solution steps  as well as just see the results (though this option isn't offered for the cubic solution above I notice!).

How do you solve the cubic equations like 8x^3+4x^2-4x-1=0 and y^3+4y^2+3y-1=0 withour a calculator? 

I used the < sign in my reply above.  I should, of course, have used the ≤ sign!

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This can be written as 2³<2^{4(7x + 2)} so that we must have 3 < 4(7x + 2)  

or 3 < 28x + 8 or -5 < 28x or -5/28 < x   or -0.1786... < x

see image below:

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