A problem regarding the hands of a clock
The easy way:
You don't actually need to find the two times when the hour and minute hand are at right angles. Just note that the difference between these two times is how long it takes for the minute hand to move $180^\circ$ farther than the hour hand has moved. The minute hand moves at $6^\circ$ per minute, and the hour hand moves at $\frac{360}{12\cdot 60}^\circ=\frac{1}{2}^\circ$ per minute. So the time required for the minute hand to advance $180^\circ$ farther than the hour hand is $\frac{180}{6-1/2}=\frac{360}{11}$ minutes.
Note that the minute hand and the hour hand are both moving with a constant angular velocity. At $12\text{pm}$, the angle between them is zero, and then after some amount of time $t$, the angle between them is $90^{\circ}$. It follows that at $2t$ the angle between them will be $180^{\circ}$; at $3t$ the angle between them will be $270^{\circ}$; and at $4t$ the minute hand and hour hand will meet again. We want to find $t$ and $3t$.
Now note that the minute hand and the hour hand meet once every hour between $1\text{pm}$ and $11\text{pm}$, and at $12\text{am}$, the minute hand and hour hand meet again, for the $11^{th}$ time. It follows that
$$11\cdot4t=12\text{ hours.}$$
Hence:
$$\begin{align*} t &= \frac{3}{11}\text{ hours}=16+\frac{4}{11}\text{ minutes} \\ \\ 3t &= \frac{9}{11}\text{ hours}=49+\frac{1}{11}\text{ minutes} \end{align*}$$
So the first time is between $12$:$16\text{pm}$ and $12$:$17\text{pm}$, and the second time is between $12$:$49\text{pm}$ and $12$:$50\text{pm}$. The difference between the two times is
$$3t-t=2t=\frac{6}{11}\text{ hours}=32+\frac{8}{11}\text{ minutes}.$$
32 minutes 44 seconds
Here's a simplified (i.e., probably not the best) way of looking at it assuming continuous, constant movement of the hands of the clock. One nice, simplifying thing is that the hour hand and minute hand start at the same point, let's say $(0,0)$.
Let's use $(t,\theta)$ to denote the internal angle of the circle $\theta$ swept out over time $t$, where $t$ is measured in minutes. For the hour hand, we start at $(0,0)$ and end at $(60,\frac{1}{12}\cdot2\pi)$ over the course of an hour. We can let the function $\theta_h(t)=\frac{\pi t}{360}$ describe such motion. Similarly, for the minute hand, we start at $(0,0)$ and end at $(60,2\pi)$ over the course of an hour. We can let the function $\theta_m(t)=\frac{\pi t}{30}$ model this motion.
Hence, we are essentially looking for a $t$ for which $\theta_m-\theta_h=\frac{\pi}{2}$ as well as a $t$ for which $\theta_m-\theta_h=\frac{3\pi}{2}$. Or, more helpfully, we are trying to find a $t$ for which $$ \frac{\pi t}{30}-\frac{\pi t}{360}=\frac{\pi}{2},\frac{3\pi}{2}, $$ and doing some basic algebra shows that what we get are solutions when $t=\frac{180}{11}, \frac{540}{11}$; hence, the difference in minutes between these two instances is $$ \frac{540}{11}-\frac{180}{11}=\frac{360}{11}=32\frac{8}{11}\;\text{minutes} $$
or, more simply, 32 minutes and 44 seconds.
Side note: The hour and minute hands will be at right angles at (roughly) 12:16:22pm and 12:49:05pm.
As said above, this is probably not the cleanest way of approaching it, but maybe it will be somewhat clearer.