Quintic Polynomial Equation

Amazingly (for a quintic), this can be solved exactly. We start with the identity $\sinh 5t=5\sinh t+20\sinh^3 t + 16\sinh^5 t$

Doubling both sides, $2\sinh 5t=10\sinh t+40\sinh^3 t + 32\sinh^5 t$

Now let $x=2\sinh t$ : $2\sinh 5t=5x+5x^3 +x^5$

Comparing this to the equation in the question, we can see we need $2\sinh 5t=1$ , so that $t=\frac15\sinh^{-1} \frac12$ and $x=2\sinh \left( \frac15\sinh^{-1} \frac12\right)\approx \boxed{0.19278}$

Even more nicely, this can be written as $x=\phi^\frac15-\phi^{-\frac15}$ where $\phi=\frac12 \left(1+\sqrt5\right)$ is the golden ratio.

If you do not know hyperbolic functions (yet), substitute $x = t - t^{-1},\:t>0$ to get a quadratic in $t^5$ . As $t>0$ we only need its positive solution: $\begin{aligned} x &= t-t^{-1}: &&&&& t^5-t^{-5}-1&=0 &&& \Rightarrow &&&& t^5&=\frac{1+\sqrt{5}}{2}=\phi &&&\Rightarrow &&&& x&=t - t^{-1}\approx\boxed{0.1928} \end{aligned}$