We want to write x^2 + bx + c = 0 in the form (x+m)^2 + n = 0, so there's only one x left and we can rearrange for it.
Expanding, (x+m)^2 = x^2 + 2mx + m^2, so we get the x^2 we want, and the coefficients of x tell us b = 2m, so m = b/2. We also get an m^2 (= b^2/4) we don't want, so let's take it away:
(x + b/2)^2 - b^2/4 = x^2 + bx
That x + b/2 is x - (-b/2), x minus the average of the roots, which is the x value the parabola is centred on. Then we add c:
(x + b/2)^2 - b^2/4 + c = x^2 + bx + c
To find the roots, set it to 0 and rearrange for the one x that's left:
(x + b/2)^2 - b^2/4 + c = 0
(x + b/2)^2 = b^2/4 - c
x + b/2 = ±√(b^2/4 - c)
x = -b/2 ± √(b^2/4 - c)
Note that this is the average of the roots ± the article's z. Then combine:
x = -b/2 ± √((b^2-4c)/4)
x = -b/2 ± √(b^2-4c)/2
x = (-b ± √(b^2-4c))/2
If you have ax^2+bx+c = 0, divide the equation by a first, then do the same steps and you get the normal quadratic formula:
x = (-b ± √(b^2-4ac))/(2a)
I think the linked post misstates the purpose of the article: it's not about new maths, but about pedagogy and ways of explaining the quadratic formula.