Solve for x using the quadratic formula in the equation: \( 3x^{2} - 6x + 2 = 0 \)

To solve the quadratic equation \( 3x^{2} - 6x + 2 = 0 \) using the quadratic formula, follow these steps: ### **Quadratic Formula Recap** For any quadratic equation of the form \( ax^{2} + bx + c = 0 \), the solutions for \( x \) are given by: \[ x = \frac{ -b \pm \sqrt{b^{2} - 4ac} }{2a} \] ### **Given Equation Analysis** Identify the coefficients from the equation \( 3x^{2} - 6x + 2 = 0 \): - \( a = 3 \) - \( b = -6 \) - \( c = 2 \) ### **Plugging into the Formula** Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula: \[ x = \frac{ -(-6) \pm \sqrt{(-6)^{2} - 4 \times 3 \times 2} }{2 \times 3} \] ### **Simplifying Step-by-Step** 1. **Calculate the Discriminant (\( \Delta \))**: \[ \Delta = b^{2} - 4ac = (-6)^{2} - 4 \times 3 \times 2 = 36 - 24 = 12 \] 2. **Compute the Square Root of the Discriminant**: \[ \sqrt{\Delta} = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \] 3. **Substitute Back into the Formula**: \[ x = \frac{6 \pm 2\sqrt{3}}{6} \] 4. **Simplify the Fraction**: \[ x = \frac{6}{6} \pm \frac{2\sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3} \] ### **Final Solutions** \[ x = 1 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 1 - \frac{\sqrt{3}}{3} \] ### **Approximate Decimal Values** For a numerical approximation: - \( \sqrt{3} \approx 1.732 \) - \( \frac{\sqrt{3}}{3} \approx 0.577 \) Thus, the solutions are approximately: \[ x \approx 1 + 0.577 = 1.577 \quad \text{and} \quad x \approx 1 - 0.577 = 0.423 \] ### **Conclusion** The quadratic equation \( 3x^{2} - 6x + 2 = 0 \) has two real solutions: \[ x = 1 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 1 - \frac{\sqrt{3}}{3} \] Or approximately: \[ x \approx 1.577 \quad \text{and} \quad x \approx 0.423 \]