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.
 A: To elaborate on the other comments, when you get an equation of the form: 
$x=bm^2 +am$
You can rewrite this as: 
$bm^2+am-x=0$
This is a quadratic equation in m. You might recall that, in general, if you have a function of the form: 
$y=Ax^2 + Bx +C$
And you want to solve for $y=0$, 
You can use the formula: 
$x=\frac{-B±\sqrt{B^2 - 4AC}}{2A}$
You can find a proof of this (if you are curious) here. 
So, here, instead of solving for x you are solving for m, and your equation looks like this: 
$y=Am^2 + Bm+C$
where the coefficients are: 
$A=b$
$B=a$
$C=-x$
Now, just substitute the coefficients into the formula. We get that: 
$m=\frac{-a±\sqrt{a^2 - 4b(-x)}}{2b}$
Which simplifies to:
$m=\frac{-a±\sqrt{a^2 + 4bx}}{2b}$
This is your answer. It is fully algebraically simplified, and unless you have values for a, b, or x, it cannot be simplified further.
A: 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.
A: For General a,b: $am+bm^2=b(m+\frac{a}{2b})^2-\frac{ba^2}{4b^2}$. Then use square roots.
