How many nonreal complex roots does the polynomial f of x equals 6 x squared minus 12 x plus 6 have?

The polynomial f(x) = 6x^2 - 12x + 6 has no nonreal complex roots. This is because the discriminant of the quadratic equation, Δ = b^2 - 4ac = (-12)^2 - 4(6)(6) = 144 - 144 = 0, is equal to 0. When the discriminant is equal to 0, the quadratic equation has two equal real roots.