Quadratic solution theorem In "How To Prove It", Ch 3.1 exercise 2 Vellman writes: 
Theorem Suppose $b^2 > 4ac$. Then the quadratic equation $ax^2+bx+c=0$ has exactly two  real distinct solutions.
Vellman then asks: To give an instance of this theorem, we replace $a,b,c$ but not $x$. Why ?
Answers online pointed that $a,b,c$ are free variables but $x$ is a bounded one.
Now what confuses me here is that $a,b,c$ in this context are arbitrary constants not free variables also if we restate the theorem as
Theorem For every real numbers $a,b $ and $ c$, if $b^2>4ac$  the quadratic equation $ax^2+bx+c=0$ has exactly two real distinct solutions.
Then $a,b,c$ become bound variables while it is still the exact same theorem.
What exactly is $x$. Is it free, bound or neither ? "CONTINUATION"
Also the theorem can be stated as:
Theorem Let $a,b$ and $c$ be real numbers and $b^2>4ac$. There exists exactly two  real distinct numbers $h_1$ and $h_2$ such that the $ah_1^2 +bh_1+c=0$ and $ah_2^2 +bh_2+c=0$.
In the last restatement $x$ was completely removed. 
What exactly is $x$ and does this sort of variable occur in other theorems?
The part after continuation apparently wasn't posted at first.
 A: The numbers $a,b$, and $c$ determine whether we have an instance of the theorem, namely whether the assumption of the theorem, $b^2>4ac$, is satisfied. Once we replace those three by specific numbers, we either have an instance ie. $b^2>4ac$ or not ie. $b^2\leq 4ac$. So different values $a,b$, and $c$ produce different instances. 
On the other hand, $x$ plays no role in the assumptions of the theorem, but in the implications of it.
Note also that we distinguish between $x$ as being the independent variable in the expression $ax^2+bx+c$, or the unknown of the equation $ax^2+bx+c=0$, or $x$ taking on specific values in order to either satisfy or not satisfy this equation. The fact that $x=h_1$ and $x=h_2$ satisfies $ax^2+bx+c=0$ does not mean that $x$ is no longer a variable. It only means that when $x$ assumes these two specific values, it will make the statement $ax^2+bx+c=0$ true. For all other values of $x$ the statement will be false. So let us not confuse the variable $x$ with specific values of it $h_1,h_2$. A variable and its values are different concepts.
A: You can rewrite your theorem as
$$\forall a,b,c,~b^2 > a c~\Rightarrow \mathrm{Card}\{x~|~ax^2 +bx+c=0\}=2$$
It starts with $\forall a,b,c$, so you can instantiate $a,b,c$ to instantiate your theorem. While there is no $\forall x$, so you cannot "choose" $x$.
A: "Free", "arbitrary" and "constant", in mathspeak, are merely figures of speech.  All Mathematics can be written without every using them.  For example, the first theorem can be rewritten as:
Suppose three real numbers, $a$, $b$ and $c$, satisfy $b^2>ac$, then there are two different real numbers $x_1$ and $x_2$ such that $ax_i^2+bx_i+c=0$ for both $i=1,2$.
(The second is wrong, unless you look for solutions in the complex field and you count multiplicities, but you probably want to think about this later.)
