I have read the paper and it gives a good educational context from which the quadratic equation may be looked at. I have also worked out simpler way to solve quadratic equations using function evaluation without much memorization of symbolic formula:
for the equation : f(x) = ax^2 +bx + c = 0
X = Z +- Sqrt [-f(Z)/a] ; Z = -b/2a
for the equation f(x) = x^2 +bx+c = 0 it is even a bit simpler:
X = Z +- Sqrt[ -f(z) ] where Z = -b/2.
Example 1: f(x) = 3x^2 -8x-35 = 0
Z = -b/2a = - (-8)/(2.3) = 4/3
F(Z) = -121/3
X = 4/3 + or - Sqrt ( -(1/3)*(-121/3) = 4/3 + or - 11/3 = {5, -7/3)
Example 2: (Simpler form): f(x) = x^2 - 4x+ 3 = 0
Z = -b/2 = -(-4)/2 = 2
f(Z) = 2^2 - 4.2 + 3 = -1
X = 2 + or - Sqrt( -(-1) = 2 +- 1 = { 3, 1}
For detail of this method, please see the following pre-print
for the equation : f(x) = ax^2 +bx + c = 0
X = Z +- Sqrt [-f(Z)/a] ; Z = -b/2a
for the equation f(x) = x^2 +bx+c = 0 it is even a bit simpler:
X = Z +- Sqrt[ -f(z) ] where Z = -b/2.
Example 1: f(x) = 3x^2 -8x-35 = 0
Z = -b/2a = - (-8)/(2.3) = 4/3
F(Z) = -121/3
X = 4/3 + or - Sqrt ( -(1/3)*(-121/3) = 4/3 + or - 11/3 = {5, -7/3)
Example 2: (Simpler form): f(x) = x^2 - 4x+ 3 = 0
Z = -b/2 = -(-4)/2 = 2
f(Z) = 2^2 - 4.2 + 3 = -1
X = 2 + or - Sqrt( -(-1) = 2 +- 1 = { 3, 1}
For detail of this method, please see the following pre-print
https://www.researchgate.net/publication/337829551_A_simple_...