How to solve for x-intercepts of...?

Descartes' rule of signs tells you there is one positive real root (x-intercept) and 0 or 2 negative ones. The rational root theorem tells you that rational roots, if any, will be ±3, ±2 or ±1, that is, factors of -6. Trying these values for x reveals none of them is a root, and it tells you the roots lie between -2 and -3, between 0 and -1, and between 3 and 4.

Just as there is a quadratic formula for solving quadratic equations, there is a cubic formula for solving cubic equations. It will give you the roots as

x = 3^(2/3)/(1 + I √2)^(1/3) + (3 (1 + I √2))^(1/3) ... ≈ 3.28995

x = -(1/2) (3 (1 + I √2))^(1/3) (1 - I √3) - (3^(2/3) (1 + I √3))/(2 (1 + I √2)^(1/3)) ... ≈ -2.58423

x = -((3^(2/3) (1 - I √3))/(2 (1 + I √2)^(1/3))) - 1/2 (3 (1 + I √2))^(1/3) (1 + I √3) ... ≈ -0.70572

No calculator is required for you to write the exact form of these roots (once you have the formula), however, one is needed if you want to evaluate them. Of course I = √(-1).

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You can use Newton's iteration method, evaluating the formula by hand. For one or two iterations, it is tedious, but not difficult. The formula will give you

.. x_next = x - (x(x^2-9)-6)/(3(x^2-3))

Evaluating this for one iteration requires 3 multiplications, one division, and 4 subtractions. Starting from guesses of -5/2 or -1/2 or 7/2, you can get to 3 or 4 decimal places of accuracy in two iterations. Altogether, that's 18 multiplications and 6 divisions for the three roots. Doing it by hand can be a lot of work.