\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \]

["Solving Quadratic Equations: Step-by-Step Guide Using the Quadratic Formula", "Understanding how to solve quadratic equations is a fundamental skill in algebra, with wide-ranging applications in science, engineering, and everyday problem-solving. One powerful tool for solving quadratics is the quadratic formula, which efficiently finds the roots of any equation in the standard form:", "[\nax^2 + bx + c = 0\n]", "In this article, we’ll walk through solving a classic quadratic equation step-by-step, starting with the formula and its derivation, followed by clear explanations and real-world relevance.", "---", "### The Quadratic Formula: From Equation to Solution", "For any quadratic equation:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "But sometimes, it’s useful to recognize the expression inside the square root—called the discriminant—as:", "[\n\Delta = b^2 - 4ac\n]", "This value determines the nature of the solutions:", "- If ( \Delta > 0 ): two distinct real roots\n- If ( \Delta = 0 ): one real double root\n- If ( \Delta < 0 ): two complex conjugate roots", "---", "### Step-by-Step Solution Example", "Consider the equation:\n[\nx = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}\n]", "1. Identify coefficients:\n From the standard form ( ax^2 + bx + c = 0 ):\n ( a = 1 ), ( b = 2 ), ( c = -8 )", "2. Compute the discriminant:\n [\n \Delta = b^2 - 4ac = (2)^2 - 4(1)(-8) = 4 + 32 = 36\n ]", "3. Apply the quadratic formula:\n [\n x = \frac{2 \pm \sqrt{36}}{2(1)} = \frac{2 \pm 6}{2}\n ]", "4. Simplify both roots:\n - First solution: ( x = \frac{2 + 6}{2} = \frac{8}{2} = 4 )\n - Second solution: ( x = \frac{2 - 6}{2} = \frac{-4}{2} = -2 )", "Thus, the two solutions are ( x = 4 ) and ( x = -2 ).", "---", "### Why This Formula Matters", "The quadratic formula reduces complex problem-solving into a straightforward algebraic process. It guarantees real or complex solutions and works for all quadratic equations—even when factoring becomes difficult.", "Key Applications:\n- Physics: modeling projectile motion\n- Engineering: optimizing structural designs\n- Economics: analyzing profit and cost curves\n- Geometry: finding intersection points of parabolas and lines", "---", "### Final Thoughts", "Mastering the quadratic formula empowers students and professionals alike to tackle quadratic problems with confidence. By understanding how coefficients shape the discriminant and the final roots, you gain deep insight into the equation’s behavior.", "Next time you encounter:\n[\nx = \frac{2 \pm \sqrt{\ ext{some expression}}}{2}\n]\nrecall this step-by-step method—your solution is just a few precise steps away.", "---", "Key search terms: quadratic formula, solve quadratic equations, discriminant, solve (x = \frac{2 \pm \sqrt{\dots}}{2}), step-by-step quadratic solutions."]