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

["# Solving Quadratic Equations: Step-by-Step Explanation of ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )", "When learning algebra, one of the most essential tools for solving quadratic equations is the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "This formula provides exact solutions for any quadratic equation in the standard form:", "[\nax^2 + bx + c = 0\n]", "In this article, we’ll break down the calculation behind the formula using a specific example: ( x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{2 \ imes 2} ), and show how to derive the solutions step-by-step.", "---", "## Understanding the Quadratic Formula Behind ( x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{2 \ imes 2} )", "The quadratic equation begins as:", "[\n2x^2 + (-8)x + 6 = 0\n]", "Here, comparing with ( ax^2 + bx + c = 0 ):", "- ( a = 2 )\n- ( b = -8 )\n- ( c = 6 )", "The formula includes three key components: ( -b ), the discriminant ( \sqrt{b^2 - 4ac} ), and ( 2a ) — these determine the nature and number of solutions.", "---", "## Step 1: Input Values into the Formula", "We substitute ( a = 2 ), ( b = -8 ), ( c = 6 ) into the quadratic formula:", "[\nx = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \ imes 2 \ imes 6}}{2 \ imes 2}\n]", "Simplify the numerator and denominator:", "- ( -(-8) = +8 )\n- Denominator: ( 2 \ imes 2 = 4 )\n- Inside the square root: compute step-by-step below", "So expression becomes:", "[\nx = \frac{8 \pm \sqrt{64 - 48}}{4}\n]", "---", "## Step 2: Compute the Discriminant", "The expression under the square root is critical — the discriminant ( D = b^2 - 4ac ):", "[\n(-8)^2 - 4 \ imes 2 \ imes 6 = 64 - 48 = 16\n]", "Since ( D = 16 > 0 ), we have two distinct real solutions (one positive, one negative due to the ( \pm ) signs).", "---", "## Step 3: Simplify the Full Expression", "Now substitute back:", "[\nx = \frac{8 \pm \sqrt{16}}{4}\n]", "Since ( \sqrt{16} = 4 ), this becomes:", "[\nx = \frac{8 \pm 4}{4}\n]", "Split into two solutions:", "1. ( x = \frac{8 + 4}{4} = \frac{12}{4} = 3 )\n2. ( x = \frac{8 - 4}{4} = \frac{4}{4} = 1 )", "---", "## Final Answer: Two Real Solutions", "The solutions to ( 2x^2 - 8x + 6 = 0 ) are:", "[\n\boxed{x = 1 \quad} \ ext{and} \quad x = 3\n]", "---", "## Why This Formula Matters", "The quadratic formula guarantees a solution (or specifies whether none, one, or two real solutions exist) without needing to factor complex quadratics. Mastering this allows students and professionals to confidently solve any quadratic equation, whether in physics, engineering, economics, or advanced mathematics.", "---", "## Key Takeaways", "- Identify coefficients ( a ), ( b ), and ( c ) from ( ax^2 + bx + c = 0 ).\n- Compute the discriminant ( b^2 - 4ac ) to determine the nature of solutions.\n- Use the formula ( x = \frac{-b \pm \sqrt{D}}{2a} ) carefully—pay attention to signs and parentheses!\n- Simplify step-by-step for clarity and accuracy.", "Understanding and applying the quadratic formula opens doors to solving real-world problems that involve parabolic motion, optimization, and more.", "---", "Keywords: quadratic formula, solve quadratic equation, ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), algebra, discriminant, real solutions, mathematics tutorial."]