Help with a quintic polynomial $$y = 0.10 + 4.060264x - 6.226862x^2 + 48.145864x^3 - 60.928632x^4 + 49.848766x^5$$
I need to be able to solve this equation for $x$.
I've looked around and seem to be failing miserably and solving this myself.
I'll have a $y$ value (likey between 0-35) and I need to find an exact $x$ (likely between 0-1).
Thanks
 A: One way to obtain a "symbolic version," as requested in the comments, is to compute some relatively simple approximation and polish it with a Newton-Raphson step.  Because this function is smooth and monotonic for $0 \le y \le 35$ this is going to work very well.
In fact, a least-squares fit of the functional form $a \log(b + c(y+1)^{1/5} + d(y-e)^2$ to the solutions for $y=0, 1, \ldots, 35$ already gets close: most of the errors are less than 0.0003 .  One Newton-Raphson step is a rational function of this expression of degree 5 (numerator) and 4 (denominator), thereby expressible in terms of 11 parameters derived from the original polynomial.  The residuals of this 16-parameter expression range from $-6 10^{-6}$ to $2.7 10^{-7}$, which is close to the precision of the original polynomial coefficients.  For $y \ge 4$ the errors are all less than $10^{-7}$, which is as good as one can hope for.
To find this solution in Mathematica, begin by generating the array of solutions for  $y=0, 1, \ldots, 35$:
Clear[x, y];
roots = x /. Table[FindRoot[-y + 0.10 + 4.060264 x - 6.226862 x^2 + 
 48.145864 x^3 - 60.928632 x^4 + 49.848766 x^5, {x, .5}], {y, 0, 35}]

Fit the initial simple model (using some eyeball guesses for the parameters):
Clear[a, b, c]; 
model = a Log[b + c  y^(1/5)] +  d (y - e)^2; 
fit = FindFit[roots, model, {{a, .5}, {b, 1}, {c, .1}, {e, 18}, {d, .0001}}, y]

Create a Newton-Raphson step for a function f at the argument a:
Clear[nr];
nr[f_, a_] := (x - f[x]/D[f[x], x]) /. x -> a

Use it to improve the model:
Clear[x];
x[z_] := ( nr[f[#] - y + 1 &, model /. fit ]) /. y -> (z + 1)

(The shift to y-1 from y is needed because Mathematica starts indexing at 1, not 0.)  The model works well for $1 \le y \le 35$ and exceptionally well for $y \ge 4$.
g = Table[x[y], {y, 1, 36}];
ListPlot[roots - g, PlotRange -> {Full, Full},
    PlotStyle -> PointSize[0.015], DataRange -> {0, 35}, 
    AxesLabel -> {"y", "Error"}]


If you need better solutions for $y \lt 4$, you could similarly fit a simple model plus a Newton-Raphson polish to this range of values alone.
A: It is a quintic, and so must have at least one real root.  The coefficient of $x^5$ is positive so $y$ is negative if $x$ is large and negative and $y$ is positive if $x$ is large and positive.
"By inspection" you will get negative $y$ for $x=-2$ and you obviously get a positive $y$ for $x=0$.  In fact it passes through the points  $(-2, -2988.113512)$ and $(0, 0.1)$, so there is a root between them so try halfway, finding the point $(-1,-169.110388)$ and continue halving the interval, so next $(-0.5,-14.8708939375)$; you don't need such precision, and in fact only need the sign of $y$.  Just keep halving between two values of $x$ which give opposite signs for $y$ until you get an answer which is precise enough for you.
There are faster root-finding algorithms, but this bisection method will work whenever you have a continuous function and two points with $y$ having opposite signs.
