Is $y^2+f(y)b+c$ a quadratic equation? The solution to the question:

Let $x, y, z \in R$ such that $x+y+z=6$ and $x y+y z+z x=7$. Then find the range of values of $x, y$, and $z$.

given in book is as follows:

$x, y, z \in R$
$x+y+z=6$ and $x y+y z+z x=7$
$\Rightarrow y(6-y-z)+y z+z(6-y-z)=7$
$\Rightarrow \quad-y^{2}+(6-z+z-z) y+z(6-z)-7=0$
$\Rightarrow \quad y^{2}+(z-6) y+7+z(z-6)=0$
.
Now, $y$ is real. Therefore,
$(z-6)^{2}-4[7+z(z-6)] \geq 0$...(1)
or $3 z^{2}-12 z-8 \leq 0$
or $\frac{12-\sqrt{144+96}}{6} \leq z \leq \frac{12+\sqrt{144+96}}{6}$
or $\frac{6-2 \sqrt{15}}{3} \leq z \leq \frac{6+2 \sqrt{15}}{3}$.
From symmetry, $x$ and $y$ have same range.

But I doubt if this is correct; since $y$ is function of $z$, hence the author cannot solve the question this way. Let me explain my point clearly.
Quadratic equations are defined as those which can be expressed in the form $ax^2+bx+c=0$, where $a \not=0$ and $a, b, c$ are constants. But clearly in this case $z$ is not constant, it is function of $y$, hence $(1)$ is not a quadratic equation so we cant the do the steps after $(1)$ hence the solution is wrong.
So the question is, is $y^2+f(y)b+c$ a quadratic equation?
 A: It sounds like you are asking: If you have a quadratic in the variable $y$, does the quadratic formula hold, if the coefficients of the quadratic depend on some other variable $z$?
In your case, if
$$y^2 + (z-6)y + (7+z(z-6)) = 0,$$
can you still treat the coefficients $z-6$ and $7+z(z-6)$ as constant and use the usual quadratic formula to find an expression for $y$? The answer is yes. If you go through the derivation of the quadratic formula, you will see that it is not important that the coefficients are constants: They are allowed to vary with other variables, and the key is that the coefficients are constant with respect to y.
So if you have an expression of the form
$$a(z)y^2 + b(z)y + c(z) = 0,$$
where $a$, $b$, and $c$ are functions of $z$, then I would indeed call this a quadratic in $y$, and I would gladly apply the quadratic formula to find an expression for $y$ in terms of $z$ (and here you of course need to take care if $a(z)\neq 0$).
A: First, the quadratic formula deals with polynomials.  If you introduce some unspecified function $f$, you muddy the waters, since $f$ needn't be a polynomial.  For example, if $f(x)=\sin x$, then $x^2+f(x)$ is not quadratic in $x$.
I understand that you saw the quadratic formula derived, it was assumed that $a,b,c$ are constants, but the proof works so long as they aren't functions of $x$.
When we say that the roots of $p(x)=ax^2+bx+c$ are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ this is true so long as $a\neq0$ and $a,b,$ and $c$ do not depend on $x$.  You can verify this by substituting them into the formula for $p$.
It doesn't matter what $a,b,$ and $c$ are, so long as they don't depend on $x$.  They could be trigonometric functions of some other variable, for example.
A: Just go back and recall the derivation of the quadratic formula, it never uses the fact that $a$, $b$ and $c$ are constant, you can use it any time. Just as an example, consider:
$$\begin{align*}
2x+4&= 20\\
4+2x-20&= 0\\
(1)\color{blue}{2}^2+(x)\color{blue}{2}+(-20)&= 0\\
\color{blue}{2}&= \dfrac{-x\pm\sqrt{x^2-4(1)(-20)}}{2(1)}\\
4+x&=\pm\sqrt{x^2+80}\\
16+8x+x^2&=x^2+80\\
x&= 8
\end{align*}$$
Here, the coefficient was itself a variable and the variable was a constant, but the quadratic formula, as you see works. Note the coefficient of variable squared must be non-zero (here, $1\neq 0$).
$x^2+f(x)b+c=0$ may or may not be a quadratic equation, but you can always use the quadratic formula!
Hope this helps. Ask anything if not clear :)
