It's the same thing. What Kline is doing is what's in general known as a depression, which is a variable substitution done for a polynomial of degree $n$, such that the resulting polynomial has no term of degree $x^{n-1}$.
One way to see how depression works is to consider the Vieta formulae, which in the case of the quadratic $ax^2+bx+c=a(x-x_1)(x-x_2)$ looks like this:
$$\begin{align*}
x_1 x_2 &= \frac{c}{a}\\
x_1 + x_2 &= -\frac{b}{a}
\end{align*}$$
From this, if the sum of the roots is $-\dfrac{b}{a}$, then the mean of the roots is $-\dfrac{b}{2a}$. Thus, the depression substitution
$$y=x+\frac{b}{2a}$$
can be geometrically interpreted as shifting the parabola corresponding to your quadratic $ax^2+bx+c$ such that it is "centered" about the origin, and the two roots are laid out symmetrically.
If we depress our original quadratic, we get
$$\require{cancel}\begin{align*}
ax^2+bx+c&=a\left(y-\frac{b}{2a}\right)^2+b\left(y-\frac{b}{2a}\right)+c\\
&=a\left(y^2-\frac{b}{a}y+\frac{b^2}{4a^2}\right)+b\left(y-\frac{b}{2a}\right)+c\\
&=ay^2\cancel{-by}+\frac{b^2}{4a}\cancel{+by}-\frac{b^2}{2a}+c\\
&=ay^2+\frac{b^2}{4a}-\frac{2b^2}{4a}+c\\
&=ay^2-\frac{b^2}{4a}+c=ay^2-\frac{b^2-4ac}{4a}
\end{align*}$$
and I'm sure you know how easy it is to solve the equation
$$ay^2-\frac{b^2-4ac}{4a}=0$$
Having solved for $y$, you undo the depression you did, which means you have to add the term $-\dfrac{b}{2a}$ to get the actual roots you want. That's where that part of the quadratic formula comes from.
In short, I would not say that Kline's method is the most expedient, but it at least looks to me that this slower method allows for more cogitation on what the steps are supposed to "mean", as opposed to a lazy plug-and-chug.
Some everyday concepts could help. Such as
In a restaurant menu (f=food item, p=price of item):
Is f a function of p? Is p a function of f?
On back of a mailed envelop (s=street address, z=5-digit zip code):
Is s a function of z? Is z a function of s?
In a teacher's grade book (n=name of student who took a test, g= grade of student)
Is n a function of g? Is g a function of n?
Best Answer
Well, a circle can be described by a function, just not in the sense that you may be familiar with. If you are looking at a function that describes a set of points in Cartesian space by mapping each $x$-coordinate to a $y$-coordinate, then a circle cannot be described by a function because it fails what is known in High School as the vertical line test.
A function, by definition, has a unique output for every input. However, for almost all points on a circle, there is another point with the same $x$-coordinate. So, you would need your function to give two different $y$-coordinates for certain inputs, which is not allowed.
However, there is no rule that the input of a function has to be an $x$-coordinate or that the output has to be a $y$-coordinate, so we can define other functions that describle a circle. In more formal terms, the domain and codomain of a function do not have to be $\Bbb{R}$. For example, we can have a function that outputs an ordered pair (that is, codomain of $\Bbb{R}\times\Bbb{R}$). Then, $$f(t)=(\sin t,\cos t)$$ outputs the unit circle when $0\le t<2\pi$. We could also describe the points in space in a different way, using polar coordinates. Here we use the counter-clockwise angle from the positive $x$-axis, $\theta$, and the distance from the origin, $r$, to identify a point. Using this system, we can easily describe the unit circle as $(\theta,f(\theta))$, where $f(\theta)=1$ and $0\le\theta<2\pi$.