what is the two factors of x^2-6x-1 x=? x=?

x^2 - 6x - 1 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)

a = 1
b = -6
c = -1

x = [6 ±√(36 + 4)]/2
x = [6 ±√40]/2
x = [6 ±√(2^2 * 10)]/2
x = [6 ±2√10]/2
x = 3 ±√10

∴ the two factors of x^2 - 6x - 1 is 3 + √10 and 3 - √10.

x² - 6x - 1 = 0
x² - 3x = 1 + 3²
x² - 3x = 1 + 9
(x - 3)² = 10
x - 3 = +/- 3.1622777

Factors:
= x - 3.1622777 - 3, = x - 6.1622777
= x + 3.1622777 - 3, = x + 0.1622777

Answer: (x - 6.1622777)(x + 0.1622777); x = 6.1622777, - 0.1622777

Proof:
= (x - 6.1622777)(x + 0.1622777)
= x² + 0.1622777x - 6.1622777x - 1
= x² - 6x - 1

(-6)^2-4(1)*(-1)=40
→X1=(-b+√∆)/2=[-(-6)+√40]/2=[6+6.32]/2=6.16
,X2=(-b-√∆)/2=[-(-6)-√40]/2=[6-6.32]/2=-0.16
or
X1=3+√40
X2=3-√40
you can check

x^2-6x-1=0
=x^2-6x+9 -9+1
=(x-3)^2-8
=(x-3+sqrt8)(x-3-sqrt8) These are the factors.

Using the null factor law, if x-3 + - sqrt8= 0
x = 3 + sqrt8 or x = 3- sqrt8

(x-3-\/'''10''')(x-3+\/''''10''')

delta=36-4.(-1)=40
x=(6-sqrt(40))/2 or x=(6+sqrt(40))/2

x=3-sqrt(10) or x=3+sqrt(10)

x= (6 (+/-) √ (36+4))/2

x= (6(+/-) √40)/2 = 3 (+/-) √10

so x= 3+√10 AND x=3-√10

aha.. dont know