How do I solve $3x^2-e^x=0$?

Question

Solve the following equation for $x$ $$3x^2-e^x=0$$

I tried introducing the logarithm, which gives $$\ln\left(3x^2\right)=x$$ but I can't see how to proceed from here.

Answer 1

Well, the are no elementary functions to solve this problem but we have that:

$$\text{n}x^2-e^x=0\space\Longleftrightarrow\space x=-2\text{W}_\text{k}\left(\pm\space\frac{1}{2\sqrt{\text{n}}}\right)\tag1$$

When $\text{n}\ne0$ and $\text{k}\in\mathbb{Z}$

Now, the real solutions are given by:

$$3x^2-e^x=0\space\Longleftrightarrow\space x=-2\text{W}\left(\pm\space\frac{1}{2\sqrt{3}}\right)\space\space\space\vee\space\space\space x=-2\text{W}_{-1}\left(-\frac{1}{2\sqrt{3}}\right)\tag2$$

Where $\text{W}\left(\text{z}\right)$ is the product log function and $\text{W}_\text{k}\left(\text{z}\right)$ is the analytic continuation of the product log function.