I am solving this equation:
$$x^2e^{0.4x}=2$$
I managed to get here:
\begin{align*} 0.2xe^{0.2x} &= \pm0.2\sqrt{2}\\ 0.2x &= W\bigl(\pm0.2\sqrt{2}\,\bigr) \end{align*}
This, however, only results in $2$ solutions, while the graph on Geogebra/Desmos clearly yields three roots. Which part of the solution am I missing? I see no reason as to where a third root would disappear during my algebraic manipulation. Thank you for your help.
You have correctly arrived at $$0.2x = W\bigl(\pm0.2\sqrt{2}\bigr),$$ which means that $$x = 5W\bigl(\pm0.2\sqrt{2}\bigr).$$ Now, $W$ has two branches, which for real numbers are $W_{0}$ and $W_{-1}$. Now, $W_{0}(x)$ is defined for $x\ge -1/e$ and $W_{-1}(x)$ is defined for $x$ in $[-1/e,0).$ Now, since $-0.2\sqrt{2} \approx -0.283 > -1/e$ we will have three solutions: \begin{align*} 5W_{0}\bigl(-0.2\sqrt{2}\,\bigr) &\approx-2.1926\\[5pt] 5W_{0}\bigl(0.2\sqrt{2}\,\bigr) &\approx 1.1285\\[5pt] 5W_{-1} \bigl(-0.2\sqrt{2}\,\bigr) &\approx -9.5497. \end{align*}