How to compare the AC Method (factoring by grouping) and the new Transforming Method in solving quadratic equations?

A good way to compare these 2 methods is solving a sample of quadratic equation.
The Transforming Method. Solve
#y = 16x^2 - 62 x + 21 = 0#
Transformed equation:
#y' = x^2 - 62x + 336 = 0# -->( ac = 336)
Proceeding: Find the 2 real roots of the transformed equation y', then, divide them by a = 16.
Find 2 numbers knowing sum (-b = 62) and product (ac = 336).
Compose factor pairs of (336) --> ...(4, 82)(6, 56). This sum is (6 + 56 = 62 = -b). Then, the 2 real roots of y' are: 6 and 56.
Back to y, the 2 real roots are:
#x1 = 6/a = 6/16 = 3/8#, and #x2 = 56/16 = 7/2#.

#y = 16x^2 - 62x + 21 - 0#
Proceed to split the middle term by proceeding as follows:
Find 2 numbers knowing sum (b = -62) and product
(ac = 16*21 = 336)
Compose factor pairs of (336):
... (-4, - 82)(-6, -56). This sum is (-62) and its product is (336).
Re-write the equation and split the middle term (-62x) into (-6x) and
(- 56x)
#y = 16x^2 - 6x - 56 x + 21#
#y = 2x(8x - 3) - 7(8x - 3)#
#y = (8x - 3)(2x - 7)#
Solve the 2 binomials:
#(8x - 3) = 0# --> #x = 3/8#
#(2x - 7) = 0# --> #x = 7/2#

NOTE. The new Transforming Method (Google Search) avoids the lengthy factoring by grouping and solving the 2 binomials.
After you find the 2 numbers (-6) and (-56). Take the opposite of them, (6) and (56), then divide them by a = 16, you immediately get the 2 real roots. You don't need to proceed factoring by grouping further.