Find X given Y in a cubic function. Having asked this question on the math overflow boards one of the contributors suggested this may be a more appropriate forum.
I have a cubic function in the form:
$$y = ax^3 + bx^2 + cx + d$$
...(where a, b, c and d are all known constants e.g -0.3, 3.5, 3.83 and 0 respectively) that produces a curve and allows me to calculate a point on that curve given it's x co-ordinate.
What I'd really appreciate help with is: How could I rewrite this equation to calculate the value of x if I knew the value of y?
I understand that cubic curves of this nature can have multiple (upto 3) values for x for any given y though I do not know how to calculate them, and my specific interest is only in the 'rightmost' x (i.e that with highest positive value).
Through research to date I gather my question relates some what to finding the roots of the cubic equation (where the given Y is specifically 0, so the values of x are where the curve crosses the horizontal axis) but I can't expand my admittedly hazy comprehension to calculating for different y values.
The only additional hint I've been able to glean is that because in my circumstances d is always 0 that I may not be dealing with a 'true' cubic function at all and instead might want to "factor out an 'x' and a quadratic".
I have noted there are similar questions posted here but these deal specifically with cubic bezier curves, where the form of the equation deals with t (time) based coefficients and the answers refer to expression of the curve in terms of control points rather then the a, b, c, d form.
Give me anything but mathematics to Google and I'm usually able to work something out for myself but in the sea of coefficients and polynomials I'm afraid I'm lost.
Thank you in adavnce for any pointers, John.

As your hint says, it is easy to solve your particular equation when $y=0$ as you have $$-0.3 x^3+0.5 x^2 + 3.83x =0$$ so either $x=0$ or $-0.3 x^2+0.5 x + 3.83 =0$ and you can solve the latter as a quadratic.  
It is usually not so easy when $y \not = 0$ or in general for a cubic. Wikipedia gives the general (sometimes complex) form of the roots
$$\begin{align}
x_1 =
&-\frac{b}{3 a}\\
&-\frac{1}{3 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&-\frac{1}{3 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
x_2 =
&-\frac{b}{3 a}\\
&+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
x_3 =
&-\frac{b}{3 a}\\
&+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d+\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}\\
&+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\tfrac12\left[2 b^3-9 a b c+27 a^2 d-\sqrt{\left(2 b^3-9 a b c+27 a^2 d\right)^2-4 \left(b^2-3 a c\right)^3}\right]}
\end{align}$$
