How do you solve 3x^2 - 2x - 2 = 03x2−2x−2=0 by completing the square?

1 Answer
Nghi N
Jul 9, 2017

x = 1/3 +- sqrt(7/9)x=13±√79

Explanation:

3x^2 - 2x = 23x2−2x=2
Divide both sides by 3
x^2 - (2x)/3 = 2/3x2−2x3=23
x^2 - (2x)/3 + 1/9 = 2/3 + 1/9 = 7/9x2−2x3+19=23+19=79
(x - 1/3)^2 = 7/9(x−13)2=79
x - 1/3 = +- sqrt(7/9)x−13=±√79
x = 1/3 +- sqrt(7/9)x=13±√79

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