Using the quadratic formula \( x = rac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -6 \), and \( c = 2 \):

["# Solving Quadratic Equations Made Simple: How to Use the Quadratic Formula with ( a = 3 ), ( b = -6 ), and ( c = 2 )", "Solving quadratic equations is a fundamental skill in algebra, and the quadratic formula stands as one of the most powerful tools in your math arsenal. Whether you're working on physics, economics, or pure math, understanding how to apply the formula correctly saves time and reduces errors. In this guide, we’ll explore how to use the quadratic formula—( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} )—with specific coefficients: ( a = 3 ), ( b = -6 ), and ( c = 2 ). We’ll walk through each step methodically, making it easy to apply this formula in real-world scenarios and academic problems alike.", "## Understanding the Quadratic Formula Setup", "The standard quadratic equation in the form ( ax^2 + bx + c = 0 ) has two solutions derived from the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Where:\n- ( a ) is the coefficient of ( x^2 )\n- ( b ) is the coefficient of ( x )\n- ( c ) is the constant term", "Given values are:\n- ( a = 3 )\n- ( b = -6 )\n- ( c = 2 )", "These coefficients put the equation in standard form:\n( 3x^2 - 6x + 2 = 0 )", "Substituting into the formula:\n[\nx = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)}\n]", "## Step 1: Compute the Discriminant", "The discriminant, ( b^2 - 4ac ), determines the nature of the roots.\nCalculate:\n[\nb^2 = (-6)^2 = 36\n]\n[\n4ac = 4 \ imes 3 \ imes 2 = 24\n]\n[\n\ ext{Discriminant} = 36 - 24 = 12\n]", "Because the discriminant is positive ((12 > 0)), the equation has two distinct real solutions.", "## Step 2: Insert Values into the Formula", "Now plug in all values into the quadratic formula:\n[\nx = \frac{6 \pm \sqrt{12}}{6}\n]", "Simplify ( \sqrt{12} ):\n[\n\sqrt{12} = \sqrt{4 \ imes 3} = 2\sqrt{3} \approx 3.464\n]", "But for exact solutions, keep it as ( \sqrt{12} ):\n[\nx = \frac{6 \pm 2\sqrt{3}}{6}\n]", "## Step 3: Simplify the Expression", "Factor numerator and denominator:\n[\nx = \frac{2(3 \pm \sqrt{3})}{6} = \frac{3 \pm \sqrt{3}}{3}\n]", "Thus, the two solutions are:\n[\nx = \frac{3 + \sqrt{3}}{3} \quad \ ext{and} \quad x = \frac{3 - \sqrt{3}}{3}\n]", "You can also write these as:\n[\nx = 1 + \frac{\sqrt{3}}{3} \quad \ ext{and} \quad x = 1 - \frac{\sqrt{3}}{3}\n]", "## Why This Matters: Real-World Applications", "The quadratic formula isn’t just a theoretical tool—it applies to countless real-life problems:\n- Physics: Calculating projectile motion and motion under constant acceleration\n- Engineering: Designing parabolic structures and optimizing design parameters\n- Finance: Modeling revenue and profit functions with quadratic relationships\n- Computer Graphics: Rendering curves and solving geometric equations", "With ( a = 3 ), ( b = -6 ), and ( c = 2 ), we saw how two rational and irrational roots emerge, providing critical insight into system behavior—such as when a quadratic model predicts two intersection points or a critical value.", "## Final Answer", "The solutions to the equation ( 3x^2 - 6x + 2 = 0 ) using the quadratic formula are:", "[\nx = \frac{3 + \sqrt{3}}{3} \quad \ ext{and} \quad x = \frac{3 - \sqrt{3}}{3}\n]", "Or approximately:\n[\nx \approx 1.577 \quad \ ext{and} \quad x \approx 0.423\n]", "### Summary", "Using the quadratic formula with ( a = 3 ), ( b = -6 ), and ( c = 2 ) reveals two distinct real roots via a clean step-by-step calculation starting from substitution, computing the discriminant, simplifying radicals, and reducing the expression. Mastery of this process equips students and practitioners to solve quadratic equations efficiently across disciplines, turning complex formulas into practical solutions.", "---", "Keywords: quadratic formula, solve quadratic equations, real roots, mathematics tutorial, rational and irrational solutions, algebra practice, ( a = 3, b = -6, c = 2 ), solve ( 3x^2 - 6x + 2 = 0 ), quadratic formula steps, algebraic solutions, discriminant analysis."]