\(x = \frac{-50 \pm \sqrt{2500 + 24000}}{2} = \frac{-50 \pm \sqrt{26500}}{2}\).

["# Solving the Quadratic Equation: (x = \frac{-50 \pm \sqrt{2500 + 24000}}{2})", "Understanding how to solve quadratic equations is fundamental in algebra and forms the basis for more complex mathematical concepts. One particularly instructive example is the equation:", "[\nx = \frac{-50 \pm \sqrt{2500 + 24000}}{2}\n]", "This equation exemplifies how simplifying expressions step-by-step leads to elegant solutions in quadratic problems. Let’s explore its derivation, calculation, and interpretation in clear, SEO-optimized detail.", "---", "## Simplifying the Discriminant: From Basics to Square Root", "Start with the expression under the square root in the quadratic formula:", "[\n\sqrt{2500 + 24000}\n]", "First, compute the sum inside the radical:", "[\n2500 + 24000 = 26500\n]", "Thus, the equation becomes:", "[\nx = \frac{-50 \pm \sqrt{26500}}{2}\n]", "But why (26500)? From solving:", "[\n50^2 = 2500 \quad \ ext{and} \quad \sqrt{2500 + 24000} = \sqrt{26500}\n]", "This discriminant ((26500)) determines the nature of the solutions — whether they are real, irrational, or repeated.", "---", "## Simplifying (\sqrt{26500}): Factor to Reduce the Square Root", "To simplify (\sqrt{26500}), we look for perfect squares that divide the number:", "- (26500 = 100 \ imes 265)\n- Since (100 = 10^2) is a perfect square, factor it out:", "[\n\sqrt{26500} = \sqrt{100 \ imes 265} = \sqrt{100} \ imes \sqrt{265} = 10\sqrt{265}\n]", "Now the equation simplifies to:", "[\nx = \frac{-50 \pm 10\sqrt{265}}{2}\n]", "---", "### Final Simplified Solution:", "Factor numerator by pulling out a 10:", "[\nx = \frac{10(-5 \pm \sqrt{265})}{2} = -5 \pm 5\sqrt{265}\n]", "---", "## Why This Form Matters: Exact Solutions and Applications", "The simplified expression:", "[\nx = -5 \pm 5\sqrt{265}\n]", "represents two precise real roots. Unlike decimal approximations, this form preserves mathematical accuracy and enables exact computation in scientific, engineering, and economic modeling.", "This solution technique — simplifying radicals, factoring, and reducing expressions — is crucial for efficiently solving quadratic equations in standardized testing, calculus prep, and real-world problem-solving.", "---", "## Real-World Usage and Further Learning", "Quadratic equations and their discriminants appear frequently in:", "- Physics: modeling projectile motion\n- Economics: profit maximization\n- Engineering: structural analysis", "Mastering simplification like in this example builds a strong foundation for advanced algebra, trigonometry, and even machine learning algorithms dealing with polynomial regression.", "---", "## Conclusion", "The equation (x = \frac{-50 \pm \sqrt{26500}}{2}) reveals how algebraic expressions simplify through careful calculation and factoring. By reducing (\sqrt{26500}) to (10\sqrt{265}), we transform a complex radical into a clean, exact solution. Embrace these step-wise methods — they empower deeper understanding and practical application in math and science.", "---", "Keywords: solve quadratic equation, simplify (\sqrt{26500}), quadratic formula simplification, (-5 \pm 5\sqrt{265}), algebraic simplification, exact solutions, discriminant, educational algebra, radical computation, math tutorial, quadratic formula example."]