`3x^2 - x - 1 = 0`
`<=> 3 . (x^2 - 1/3x + 1/36) - 13/12 = 0`
`<=> 3 . [x^2 - 2 . x . 1/6 + (1/6)^2 ] - 13/12 =0`
`<=> 3 . (x - 1/6)^2 - 13/12 = 0`
`<=> 3 . (x - 1/6)^2 = 13/12`
`<=> (x-1/6)^2 = 13/12 : 3`
`<=> (x-1/6)^2 = 13/36`
`<=> (x - 1/6)^2= ((\sqrt{13})/6)^2`
`<=> x - 1/6 = (\sqrt{13})/6 ` hoặc `x-1/6 = (-\sqrt{13})/6`
`+) x - 1/6 = (\sqrt{13})/6`
`<=> x = (\sqrt{13})/6 + 1/6`
`<=> x= (1 + \sqrt{13})/6`
`+) x - 1/6 = (-\sqrt{13})/6`
`<=> x = (-sqrt{13})/6 + 1/6`
`<=> x = (1 - \sqrt{13})/6`
Vậy `x\in{ (1 + \sqrt{13})/6 ; (1 - \sqrt{13})/6}`
