2u^2 + u - 1 = 0 \implies u = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}

["# Solving the Quadratic Equation: 2u² + u – 1 = 0 Step-by-Step", "Quadratic equations form the backbone of algebra and appear frequently in science, engineering, physics, and everyday problem-solving. One common form you may encounter is:", "[\n2u^2 + u - 1 = 0\n]", "This equation may seem intimidating at first, but solving it becomes straightforward using the quadratic formula. In this article, we’ll walk through the process of solving (2u^2 + u - 1 = 0) and explain how to derive the solutions in a clear, educational way—perfect for students, educators, and math enthusiasts.", "---", "## Step 1: Understand the Standard Form", "The general form of a quadratic equation is:", "[\nau^2 + bu + c = 0\n]", "For our equation:", "[\n2u^2 + u - 1 = 0\n]", "We identify the coefficients:\n- ( a = 2 )\n- ( b = 1 )\n- ( c = -1 )", "---", "## Step 2: Apply the Quadratic Formula", "The quadratic formula provides the solutions for ( u ):", "[\nu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Substitute ( a = 2 ), ( b = 1 ), and ( c = -1 ):", "[\nu = \frac{-(1) \pm \sqrt{(1)^2 - 4(2)(-1)}}{2(2)}\n]", "Simplify step by step:", "- The numerator begins with ( -1 \pm \sqrt{1 + 8} )\n- Inside the square root: ( 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 )", "So now:", "[\nu = \frac{-1 \pm \sqrt{9}}{4}\n]", "[\nu = \frac{-1 \pm 3}{4}\n]", "---", "## Step 3: Find the Two Solutions", "Using the values:", "[\nu = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}\n]", "and", "[\nu = \frac{-1 - 3}{4} = \frac{-4}{4} = -1\n]", "Thus, the solutions are:", "[\nu = \frac{1}{2} \quad \ ext{and} \quad u = -1\n]", "---", "## Step 4: Verify the Solutions", "It’s always good practice to check your answers by substituting back into the original equation.", "- For ( u = \frac{1}{2} ):", "[\n2\left(\frac{1}{2}\right)^2 + \frac{1}{2} - 1 = 2 \cdot \frac{1}{4} + \frac{1}{2} - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0 \quad \ ext{✅}\n]", "- For ( u = -1 ):", "[\n2(-1)^2 + (-1) - 1 = 2(1) - 1 - 1 = 2 - 1 - 1 = 0 \quad \ ext{✅}\n]", "Both values satisfy the equation—our solutions are correct!", "---", "## Why This Matters: Real-World Applications", "Quadratic equations model a range of phenomena, including projectile motion, electrical circuits, and optimization problems. The ability to solve them quickly and accurately is essential for anyone in STEM fields.", "Understanding derivation, like using the quadratic formula step-by-step, builds not only problem-solving skills but also deep insight into how mathematical relationships work.", "---", "## Conclusion", "Solving ( 2u^2 + u - 1 = 0 ) using the quadratic formula reveals two real solutions:", "[\nu = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}\n]", "These yield ( u = \frac{1}{2} ) and ( u = -1 )—simple yet powerful results that illustrate the elegance and utility of algebra. Whether you're a student learning for the first time or a professional revisiting fundamentals, mastering these steps strengthens your mathematical toolkit.", "---", "### Related Keywords:\n- Solve quadratic equation\n- Use quadratic formula\n- Solve 2u² + u – 1 = 0\n- Quadratic solutions derivation\n- Algebraic problem solving\n- Quadratic formula explanation\n- Roots of quadratic equation", "---", "Optimize your quadratic equation solving skills today—understanding beginnings leads to confident applications!"]