Using the quadratic formula, \( w = \frac{-5 \pm \sqrt{25 + 1200}}{4} \).
["Using the Quadratic Formula: Solving ( w = \frac{-5 \pm \sqrt{25 + 1200}}{4} ) with Ease", "When solving quadratic equations, the quadratic formula is an essential and reliable tool. One common problem involves calculating variable ( w ) using the equation:", "[\nw = \frac{-5 \pm \sqrt{25 + 1200}}{4}\n]", "This formula appears in contexts ranging from physics to finance, and understanding how to simplify and solve it empowers students, educators, and professionals alike.", "### Understanding the Quadratic Formula", "The general form of a quadratic equation is:", "[\nax^2 + bx + c = 0\n]", "The quadratic formula solves for ( x ) as:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "In our specific case, the expression inside the square root begins as ( \sqrt{25 + 1200} ), which simplifies the discriminant, making calculations straightforward.", "### Simplifying the Discriminant", "Start by evaluating the discriminant:\n[\n25 + 1200 = 1225\n]", "Notice that ( 1225 ) is a perfect square:\n[\n\sqrt{1225} = 35\n]", "This simplification is critical—it avoids complex roots and simplifies the formula significantly.", "### Substituting Back: Final Expression for ( w )", "Substitute ( a = 1 ), ( b = -5 ), and ( \sqrt{1225} = 35 ) into the quadratic formula:", "[\nw = \frac{-(-5) \pm 35}{4} = \frac{5 \pm 35}{4}\n]", "This yields two solutions:", "[\nw = \frac{5 + 35}{4} = \frac{40}{4} = 10\n]\n[\nw = \frac{5 - 35}{4} = \frac{-30}{4} = -7.5\n]", "Thus, the solutions are:", "[\nw = 10 \quad \ ext{or} \quad w = -7.5\n]", "### Why This Formula Matters in Real-World Applications", "Solving quadratic equations like this arises naturally in many domains:", "- Engineering: Calculating trajectories and frequency responses involving quadratic motion.\n- Economics: Modeling profit and cost curves where revenue follows a quadratic trend.\n- Physics: Determining time durations or distances derived from quadratic motion equations.", "Using the quadratic formula with simplified discriminants reduces arithmetic errors and accelerates problem-solving, especially in high-stakes or time-sensitive environments.", "### Step-by-Step Summary", "1. Identify coefficients from the standard form.\n2. Compute the discriminant: ( b^2 - 4ac ).\n3. Simplify the square root when possible—here, ( \sqrt{1225} = 35 ).\n4. Substitute into the formula with ( \pm ) to find two solutions.\n5. Simplify the resulting expressions.", "### Final Thoughts", "Mastering the quadratic formula and its manipulation is foundational to algebra proficiency. Whether you're a student tackling homework or a professional using math in daily decision-making, knowing how to efficiently solve equations like ( w = \frac{-5 \pm \sqrt{25 + 1200}}{4} ) ensures accuracy and speed.", "Start with simplification—often the key to unlocking clear, confident solutions.", "---", "Keywords: quadratic formula, solving quadratics, discriminant calculation, ( w = \frac{-5 \pm \sqrt{25 + 1200}}{4} ), step-by-step solution, algebra practice, real-world applications."]
