solve the equation algebraically?

x^2 - 4x - 3 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 1
b = -4
c = -3

x = [4 ±√(16 - 48)]/2
x = [4 ±√-32]/2 (imaginary number)
(no real roots)

x² - 4x - 3 = 0
x² - 4x = 3
x² - 2x = 3 + (- 2)²
x² - 2x = 3 + 4
(x - 2)² = 7
x - 2 = 2.64575

1st value of x:
x = 2 + 2.64575
x = 4.64575

2nd value of x:
x = - 2.64575 + 2
x = - 0.64575

Answer: x = 4.64575, - 0.64575; Factors: (x - 4.64575)(x + 0.64575)

Proof (x = 4.64575):
4.64575² - 4(4.64575) = 3
21.583 - 18.583 = 3
3 = 3

Proof (x = - 0.64575):
- 0.64575² - 4(- 0.64575) = 3
0.417 + 2.583 = 3
3 = 3

x = [ 4 ± √ (16 + 12 ) ] / 2
x = [ 4 ± √ (28) ] / 2
x = [ 4 ± 2√7 ] / 2
x = 2 ± √7

Uh.. Factor it?
x^2 - 4x - 3 = 0
(x - 1)(x - 3) = 0
x = 1 or x = 3

Yeah.. I'm pretty sure..