Absolutely, let's solve this quadratic equation.
First of all, let's write the equation:
9x^2 - 6x + 2 = 0
A quadratic equation is generally depicted as ax^2 + bx + c = 0, where, a, b and c are constants.
For our equation, we have a=9, b=-6, and c=2.
Recall the quadratic formula for the solutions of the quadratic equation ax^2 + bx + c = 0. The solutions (or roots) x are given by:
x = [-b ± sqrt(b^2 - 4ac)] / 2a
Substituting the values of a, b and c:
x = [6 ± sqrt((-6)^2 - 4*9*2)] / 2*9
x = [6 ± sqrt(36 - 72)] / 18
x = [6 ± sqrt(-36)] / 18
As you can see, under the square root we have a negative number. When a negative number appears under the square root in the context of real numbers, it has no solution, but in the context of complex numbers it does.
Taking square root of -1 gives us 'i' in the imaginary number system. So, sqrt(-36) = 6i.
Plugging this value back into our equation for x, we get:
x = [6 ± 6i] / 18
x = (1/3 ± 1/3i)
Therefore, the solutions for the equation 9x^2 - 6x + 2 = 0 in the complex number system are x = 1/3 - i/3 and 1/3 + i/3.