# Solving an equation with an unknown constant?

$$\frac{1}{12}Q^2=qQ-q^2, Q \in \mathbb{R}$$

How come if I let $Q=1$ and solve for $q$ then $q$ will be exactly the same as $\frac{q}{Q}$?

As in, if I solve the above equation for real, I get $\frac{q}{Q} = 0.908$ as one possible answer. Why, if I let $Q=1$ and solve for $q$ do I get the exact same answer, $q=0.908$?

I've seen it used a lot as a common physics trick to solve equations, but I'm wondering exactly why?

Not sure I understand the question completely, but I'm hoping this will answer it. First consider the case that $q=0$. Then $Q=0$ is a solution. Otherwise, dividing both sides by $q^2$, we get
$$\frac1{12}\times\frac{Q^2}{q^2}=\frac Qq-1$$ $$\frac1{12}\left(\frac Qq\right)^2-\frac Qq-1=0$$
which is a quadratic equation with variable $\frac Qq$. If $\frac Qq=a$ is a solution, this can be rewritten as $Q=aq$. Setting $q=1$ then yields a solution of $Q=a$.
I probably should have divided by $Q^2$ given the way it's worded, but it follows the same logic.