To solve the quadratic equation \(3x^2 - 6x + 2 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In this equation, the coefficients are:
- \(a = 3\)
- \(b = -6\)
- \(c = 2\)
Now, we can calculate the discriminant (\(b^2 - 4ac\)):
\[ b^2 = (-6)^2 = 36 \] \[ 4ac = 4 \cdot 3 \cdot 2 = 24 \] \[ b^2 - 4ac = 36 - 24 = 12 \]
Since the discriminant is positive, we will have two real and distinct solutions.
Now, substituting into the quadratic formula:
\[ x = \frac{-(-6) \pm \sqrt{12}}{2 \cdot 3} \] \[ x = \frac{6 \pm \sqrt{12}}{6} \]
To simplify \(\sqrt{12}\):
\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \]
Substituting this back into our equation gives:
\[ x = \frac{6 \pm 2\sqrt{3}}{6} \]
Now simplify further:
\[ x = 1 \pm \frac{\sqrt{3}}{3} \]
So the solutions are:
\[ x = 1 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 1 - \frac{\sqrt{3}}{3} \]
These are the two solutions to the quadratic equation \(3x^2 - 6x + 2 = 0\).