Let's try to follow Kline's advice (and all the time we assume the coefficients $\,a,b,c\,$ are real).
We have $\,ax^2+bx+c=0\,\,,\,a\neq 0\,$ and we make the substitution $y:=x+\frac{b}{2}\Longleftrightarrow x = y-\frac{b}{2}$ and from here, substituting in the equation
$a\left(y-\frac{b}{2}\right)^2+b\left(y-\frac{b}{2}\right)+c=0$ ...and now? But for messing up the equation's variable I can't see any real advance or simplification.
I think what he actually meant, or should have meant, is the following, which is only the rather well-known method of completing the square (CS): $X^2\pm BX=\left(X \pm \frac{B}{2}\right)^2-\frac{B^2}{4}\,\,,\,\text{and from here}$ $0=ax^2+bx+c\,\Longrightarrow\,x^2+\frac{bx}{a}=-\frac{c}{a}\stackrel{CS}\Longrightarrow\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}=-\frac{c}{a}\,\Longrightarrow$ $\Longrightarrow\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\,\,\,\text{(see the numerator?!)}\,\,\stackrel{\text{sq.rt. in both sides}}\Longrightarrow$ $\Longrightarrow x_{1,2}+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\,\Longrightarrow\,x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .
So, as you can see, the benefits is that you need that expression to solve quadratics..as simple as that!
Of course, the number $\,\Delta:=b^2-4ac\,$ , called the quadratic's discriminant serves to know beforehand about the above equation's possible solutions: $\begin{align*}\Delta>0\Longrightarrow & \,\text{there exist two different real solutions to the equation}\\ \Delta=0\Longrightarrow & \,\text{there exists only one unique real solution to the equation}\\ \Delta<0\Longrightarrow & \,\text{the equation has no real solutions}\end{align*}$
In the last case above there exist two conjugate complex non-real solutions (disregard this if you haven't yet studied complex numbers)