How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.

Taking into account the quality of previous answers and comments, my answer will probably be stupid.
If you want $x$ such that $$f(x)=x^5 +x^3+x - y=0$$ you could just apply iteratively Newton which will give as the simplest $$x_{n+1}=\frac{4 x_n^5+2 x_n^3+y}{5 x_n^4+3 x_n^2+1}$$ which could be considered as a recursive definition of the inverse of $y$. 
Assuming $y>1$, starting with $x_0=y^{\frac{1}{5}}$ (since, for this value $f(x_0) >0$ and $f''(x_0)> 0$) the process will converge without any overshoot of the solution.
Let me try with $y=10^{10}$ so $x_0=100$; the successive iterates are $x_1=99.99799992$, $x_2=99.99799984$ which is the solution for ten significant figures.
Let me try with $y=10^{5}$ so $x_0=10$; the successive iterates are $x_1=9.979920876$, $x_2=9.979839999$ which is the solution for ten significant figures.
