which two values of x are roots of the polynomial below 3x²-3x+1​

Answer:  x  = (3 − i√3)/6 and (3 + i√3)/6Step-by-step explanation:3x² − 3x + 1 = 0 ∴ 3x² − 3x = -1 ∴ x² − x = -1/3 ∴ x² − x + (-1/2)² = (-1/2)² − 1/3 given x² + bx + (b/2)² = (x + b/2)² ∴ (x − 1/2)² = 1/4 − 1/3 ∴ (x − 1/2)² = 3/12 − 4/12 ∴ (x − 1/2)² = -1/12 ∴ x − 1/2 = ±√(-1/12) This tells us there are no real roots and if you need real number solutions we stop here ∴ x − 1/2 = ±i/(2√3) ∴ x − 1/2 = ±i√(3)/6 ∴ x  = 1/2 ± i√(3)/6 ∴ x  = 3/6 ± i√(3)/6 ∴ x  = (3 − i√3)/6 and (3 + i√3)/6 <= the 2 complex roots