Substitution Method for Quadratic?

Let x = y + k

Substitute x = y + 1 into the equation 2x² - 3x + 1 = 0 :
2(y + k)² - 3(y + k) + 1 = 0
2(y² + 2ky + k²) - 3y - 3k + 1 = 0
2y² + 4ky + 2k² - 3y - 3k + 1 = 0
2y² + (4k - 3)y = -2k² + 3k - 1 …… [1]

When k = 3/4, the coefficient of y = 0

Substitute k = 3/4 into [1] :
2y² = -2(3/4)² + 3(3/4) - 1
2y² = 1/8
y² = 1/16
y = 1/4 or y = -1/4

Since x = y + (3/4), then y = x - (3/4)
x - (3/4) = 1/4 or x - (3/4) = -1/4
x = (1/4) + (3/4) or x = -(1/4) + (3/4)
x = 1 or x = 1/2

Let`s keep it simple !
[ 2x - 1 ] [ x - 1 ] = 0
x = 1/2 , x = 1

2 (y + k)² - 3 (y + k) + 1 = 0
=> 2 y² + 4 k y + 2 k² - 3 y - 3 k + 1 = 0
=> 2 y² + (4 k - 3) y + 2 k² - 3 k + 1 = 0
4 k - 3 = 0 => k = 3/4
=> 2 y² + 0 y + 18/16 - 36/16 + 16/16 = 0
=> 2 y² = 2/16
=> y² = 1/16
=> y = +- 1/4
=> x = 3/4 +- 1/4 = 1/2 or 1

x = y + k … (a)

Solve 2x² - 3x + 1 = 0 … (1)

Equation substitute … (a)

2(y + k)² - 3(y + k) + 1 = 0

2[y² + k² + 2yk] - 3y - 3k + 1 = 0

2y² + 2k² + 4yk - 3y - 3k + 1 = 0

2y² + y(4k - 3) + (2k² - 3k + 1) = 0

Compare

ax² + bx + c = 0

a = 2 b = 4k - 3 c = 2k² - 3k + 1

y = 0

4k - 3 = 0

K = 3/4

2y²

Nature of roots ∆ = b²-4ac

(4k - 3)² - 4(2) (2k² - 3k + 1)

16k² + 9 - 24k - 16k² + 24k + 8

16k² - 24k + 9 - 16k² + 24k - 8 = 0

∆ = 1

Real roots

Replace.