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.