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let f(x) = 3x^2 - 3x + 1. The roots of f is the same as asking for x where f(x) = 0.

So we do it f(x) = 0, 3x^2 - 3x + 1 = 0. Using "complete the square" method

3(x^2 - x) + 1 = 0

3(x^2 + (2 . - 1/2 . x) + (-1/2)^2 -(-1/2)^2) + 1 = 0

3(x - 1/2)^2 -3/4 + 1 = 0

3(x - 1/2)^2 + 1/4 = 0

By the quadratic formula, the solution comes out to be

x = [3+√(-3)]/6 and x = [3-√(-3)]/6

or

x = (3+i√3)/6 and x = (3-i√3)/6

in other words, the solutions are complex numbers.

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