How to derive the $\log$ of fraction of two Gaussian distribution
Your final condition is correct, but it doesn't show explicitly for which values of $x$ it is satisfied. This can be worked out as follows. From what you have it follows that $p_1(x)\ge p_2(x)$ is equivalent to

$$0\ge x^2(\sigma_2^2-\sigma_1^2)-2x(\mu_1\sigma_2^2-\mu_2\sigma_1^2)+\mu_1^2\sigma_2^2-\mu_2^2\sigma_1^2-2\sigma_1^2\sigma_2^2\log\left(\frac{\sigma_2}{\sigma_1}\right)\tag{1}$$

Let's first consider the case $\sigma_1=\sigma_2$. From $(1)$ it follows that in this case the condition on $x$ becomes

$$x\ge \frac{\mu_1^2-\mu_2^2}{2(\mu_1-\mu_2)}
= \frac{\mu_1+\mu_2}{2}\tag{2}$$

where we assume $\mu_1\neq\mu_2$ because otherwise $p_1(x)=p_2(x)$, which makes no sense.

In the case $\sigma_1\neq\sigma_2$, we have to consider $(1)$ as a quadratic inequality in $x$, which is satisfied with equality for the following values of $x$:

$$\begin{align}x_1&amp;=\frac{a}{2}-\sqrt{\frac{a^2}{4}-b}\\
x_2&amp;=\frac{a}{2}+\sqrt{\frac{a^2}{4}-b}\end{align}\tag{3}$$

with

$$\begin{align}a&amp;=\frac{2(\mu_1\sigma_2^2-\mu_2\sigma_1^2)}{\sigma_2^2-\sigma_1^2}\\
b&amp;=\frac{\mu_1^2\sigma_2^2-\mu_2^2\sigma_1^2-2\sigma_1^2\sigma_2^2\log\left(\frac{\sigma_2}{\sigma_1}\right)}{\sigma_2^2-\sigma_1^2}\end{align}\tag{4}$$

Depending on the sign of $\sigma_2-\sigma_1$ we have two conditions equivalent to $p_1(x)\ge p_2(x)$:

$$\begin{align}\sigma_2&gt;\sigma_1:\quad &amp; x_1\le x\le x_2\\
\sigma_2&lt;\sigma_1:\quad &amp; x\le x_1\text{ or }x\ge x_2\end{align}\tag{5}$$

The inequalities in $(5)$ state explicit conditions on the variable $x$ that can be easily evaluated.

EDIT: As suggested by Dilip Sarwate in a comment it can be instructive to distinguish another special case: $\sigma_1\neq\sigma_2$ and $\mu_1=\mu_2=\mu$. In that case the limits $x_1$ and $x_2$ of the decision regions given by $(3)$ lie symmetrical to the common mean $\mu$, and $(3)$ simplifies to

$$x_{1,2}=\mu\pm\sqrt{\frac{2\sigma_1^2\sigma_2^2\log\left(\frac{\sigma_2}{\sigma_1}\right)}{\sigma_2^2-\sigma_1^2}}$$

Of course, the conditions $(5)$ remain valid.