<s>[INST] How to simplify this equation to solve for m? It has been way too many years since high school. How can I simplify this equation to solve for m:
$\frac{x}{c+pm}=m$
I got to
$x = cm + pm^2$
and I don't know how to get any further. I wish this stuff stayed in my head over the last 20 years.
 [/INST] Your equation
$$
x = c\, m + p\, m^2 \quad (1)
$$
is a quadratic equation of the form
$$
y = a \, x^2 + b \, x + c
$$
(where $x$ is the unknown, in your case $m$).
If $p \ne 0$, then we can divide both sides of equation $(1)$ by $p$ to get
$$
m^2 + \frac{c}{p} m = \frac{x}{p}
$$
we now try to align the left hand side for application of the binomic formula
$$
a^2 + 2 a b + b^2 = (a + b)^2 \quad (2)
$$
where $m$ is the $a$ and $c/p$ is $2 b$, so the $b^2$ is missing, we add this to both sides:
$$
m^2 + 2 m \frac{c}{2p} + \left(\frac{c}{2p}\right)^2 = 
\frac{x}{p} + \left(\frac{c}{2p}\right)^2 
$$
we can now apply equation $(2)$ and have a new left hand side:
$$
\left(m + \frac{c}{2p} \right)^2 = \frac{x}{p} + \left(\frac{c}{2p}\right)^2 
$$
Taking the square root of both sides gives 
$$
m + \frac{c}{2p}= \pm \sqrt{\frac{x}{p} + \left(\frac{c}{2p}\right)^2} 
$$
and solving for $m$ gives:
$$
m = -\frac{c}{2p} \pm \sqrt{\frac{x}{p} + \left(\frac{c}{2p}\right)^2} 
$$
The $\pm$ symbol indicates up to two solutions, one having the added square root, the other having the substracted square root.
To have solutions which are real numbers, the radicand 
$$
\frac{x}{p} + \left(\frac{c}{2p}\right)^2
$$
must be non-negative, otherwise you get no real numbers as solutions, but complex numbers with non-zero imaginary components.
</s>