\(x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 2 \times 6}}{4} = \frac{8 \pm \sqrt{64 - 48}}{4} = \frac{8 \pm \sqrt{16}}{4}\).

["# Solving Quadratic Equations: How to Evaluate (x = \frac{8 \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{4})", "Solving quadratic equations is a fundamental skill in algebra, essential for students, educators, and professionals alike. One common form appears frequently in mathematics: quadratic expressions in the standard form (ax^2 + bx + c = 0), and learning how to simplify and solve these expressions step-by-step makes complex problems much easier to handle.", "This article breaks down the quadratic formula applied to the specific equation:\n[\nx = \frac{8 \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{4}\n]", "We’ll guide you through simplifying the square root, reducing the expression fully, and finally determining the solutions.", "---", "## Understanding the Quadratic Formula", "The quadratic formula gives the solutions to any quadratic equation:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For the equation (ax^2 + bx + c = 0):\n- (a) is the coefficient of (x^2),\n- (b) is the coefficient of (x),\n- (c) is the constant term.", "Traditional learning often emphasizes memorizing this formula, but understanding each component deepens problem-solving confidence. In our example, the numerator contains:", "[\n8 \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}\n]", "This expression represents the discriminant, a key part that determines the nature of the roots.", "---", "## Simplifying Inside the Square Root", "Start with the phrase under the square root:\n[\n(-8)^2 - 4 \ imes 2 \ imes 6\n]", "Break it down step-by-step:", "1. Square the term:\n[\n(-8)^2 = 64\n]", "2. Multiply the coefficients in the next term:\n[\n4 \ imes 2 \ imes 6 = 48\n]", "3. Subtract the two results:\n[\n64 - 48 = 16\n]", "So, the square root simplifies cleanly:\n[\n\sqrt{(-8)^2 - 4 \ imes 2 \ imes 6} = \sqrt{16} = 4\n]", "---", "## Putting It All Together", "Now substitute this simplified value back into the original equation:\n[\nx = \frac{8 \pm \sqrt{16}}{4} = \frac{8 \pm 4}{4}\n]", "Next, split into two solutions using the ± symbol:", "- First solution (using +):\n[\nx = \frac{8 + 4}{4} = \frac{12}{4} = 3\n]", "- Second solution (using –):\n[\nx = \frac{8 - 4}{4} = \frac{4}{4} = 1\n]", "---", "## Final Answer", "The two solutions to the equation are:\n[\nx = 1 \quad \ ext{and} \quad x = 3\n]", "---", "## Why This Process Matters", "This example demonstrates that simplifying expressions like (x = \frac{8 \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{4}) is not just algebraic manipulation — it’s about:", "- Recognizing perfect squares (e.g., (\sqrt{16} = 4))\n- Applying the quadratic formula correctly and precisely\n- Avoiding arithmetic errors by simplifying step-by-step\n- Understanding how the discriminant ((b^2 - 4ac)) affects roots — here, the positive discriminant indicates two distinct real solutions", "Mastering these skills is valuable not only for academic success but also for engineering, physics, economics, and any field relying on mathematical modeling.", "---", "## Want to Practice More?", "Try applying the quadratic formula to other equations like:\n[\n\frac{5 \pm \sqrt{100 - 3 \ imes 7 \ imes 2}}{6}\n]\nor\n[\n\frac{-3 \pm \sqrt{49 - 4 \ imes (-2) \ imes (-5)}}{2 \ imes (-2)}\n]", "Use the same structured approach—simplify inside the square root first, then compute, and solve.", "---", "### Key Takeaways:\n- The discriminant determines root nature: positive = two real roots, zero = one real root, negative = complex roots.\n- Always simplify the square root expression first before performing arithmetic.\n- The ± symbol ensures both possible solutions are captured.", "Understanding this process builds a strong foundation for mastering quadratic equations and beyond.", "---", "Keywords: quadratic formula, solve quadratic equation, explain discriminant, simplify square root, algebra tutorial, quadratic solutions, step-by-step math, solve (x = \frac{8 \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{4})", "---", "References:\n- Khan Academy quadratic equations\n- Paul’s Online Math Notes\n- Algebra Fundamentals Textbook Editions", "---", "Meta Description:\nLearn how to solve (x = \frac{8 \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{4}) step-by-step, including simplifying the discriminant, applying the quadratic formula, and finding real solutions. Perfect for students mastering algebra."]