Using the quadratic formula, \( n = \frac{-1 \pm \sqrt{1 + 1680}}{2} \).

["Using the Quadratic Formula: Solve Quadratic Equations with Ease", "Solving quadratic equations is a fundamental skill in algebra, widely applicable in science, engineering, economics, and everyday problem-solving. One powerful tool for finding the roots of any quadratic equation is the quadratic formula, particularly when equations are difficult to factor. In this article, we explore a common quadratic equation written in standard form and demonstrate how to apply the quadratic formula step-by-step, including the specific expression:", "[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2}\n]", "---", "### Understanding the Quadratic Formula", "The quadratic formula solves equations of the form:\n[\nax^2 + bx + c = 0\n]\nwith the solution:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "In our given example, the quadratic equation is implicitly written as:\n[\nn^2 + n + 1680 = 0\n]\nwhich matches the standard form with:\n- ( a = 1 )\n- ( b = 1 )\n- ( c = 1680 )", "---", "### Step 1: Compute the Discriminant", "The discriminant ( D ) determines the nature of the roots and is calculated as:\n[\nD = b^2 - 4ac\n]", "Substitute the values:\n[\nD = 1^2 - 4(1)(1680) = 1 - 6720 = -6719\n]", "Wait — this result is negative, suggesting complex roots. However, the given formula shows the square root of ( 1 + 1680 ), which evaluates to:\n[\n\sqrt{1 + 1680} = \sqrt{1681}\n]", "But we know that:\n[\n\sqrt{1681} = 41\n]\nsince ( 41^2 = 1681 ).", "This discrepancy arises because the formula in the prompt appears to simplify ( b^2 + 1680 ) assuming ( b = 1 ), so indeed:\n[\n\sqrt{b^2 + 1680} = \sqrt{1 + 1680} = \sqrt{1681} = 41\n]", "Let’s correct and clarify: the full quadratic formula with accurate values is:", "[\nn = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4(1)(1680)}}{2(1)} = \frac{-1 \pm \sqrt{1 - 6720}}{2}\n]", "But the original expression in the prompt uses:\n[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2} = \frac{-1 \pm 41}{2}\n]", "This assumes the equation is:\n[\nn^2 + n + 1680 = 0 \Rightarrow \ ext{but wait, } b^2 = 1, \quad D = 1 - 6720 = -6719 \Rightarrow \ ext{contradiction}\n]", "Ah — here’s the resolution: likely, the original equation was intended as\n[\nn^2 + n - 1680 = 0 \quad \ ext{or sometimes } n^2 - n + 1680 = 0\n]\nBut the root square root ( \sqrt{1 + 1680} ) points clearly to a positive discriminant, so the equation should resemble one where:\n[\nb^2 + 1680 = (1)^2 + 1680 = 1681 = 41^2\n]", "So the intended equation is probably:\n[\nn^2 + n - 1680 = 0\n]\nor more likely:\n[\nn^2 - n + 1680 = 0 \quad \ ext{(but then } b = -1 \ ext{)}\n]", "But given the formula:\n[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2}\n]\nwe accept that the equation was likely simplified or miswritten — but mathematically, if ( b = 1 ), ( a = 1 ), then ( b^2 + 1680 = 1681 ) implies:\n[\n\sqrt{b^2 + 1680} = \sqrt{1 + 1680} = 41\n]\nso the root expression comes from:\n[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2} = \frac{-1 \pm 41}{2}\n]", "Thus, the intended quadratic equation is:\n[\nn^2 + n + 1680 = 0 \quad \ ext{was probably meant} \quad n^2 + n - 1680 = 0\n]\nor more carefully:\n[\nn^2 + n + c = 0 \ ext{ with } c \ ext{ such that } 1 - 4c = 1681 \Rightarrow -4c = 1680 \Rightarrow c = -420\n]\nBut to match the given expression exactly, we treat the equation as:\n[\nn^2 + n + 1680 = 0\n]\nwith a corrected discriminant:\nWait — ( 1 - 4(1)(1680) = -6719 ), not 1681.", "Conclusion: There is a typo in the original expression. The correct evaluation is:\n[\n\sqrt{b^2 - 4ac} = \sqrt{1 - 6720} = \sqrt{-6719}\n]\nso the roots are complex, unless the equation is misstated.", "But since the prompt specifies:\n[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2} = \frac{-1 \pm 41}{2}\n]\nwe interpret this as:\n[\n\boxed{n = \frac{-1 \pm \sqrt{1 + 1680}}{2} \Rightarrow n = \frac{-1 \pm 41}{2}}\n]\nmeaning the underlying equation was likely\n[\nn^2 + n - 420 = 0 \quad \ ext{or similar}, \ ext{ but with sign correction}\n]", "Alternatively, the expression ( \sqrt{1 + 1680} ) is shorthand for a specific problem setup where the discriminant evaluates neatly.", "---", "### Step 2: Apply the Simplified Formula", "Given the clean numbers — ( b = 1 ), ( b^2 + 1680 = 1 + 1680 = 1681 = 41^2 ):", "Use the quadratic formula:\n[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2} = \frac{-1 \pm \sqrt{1681}}{2}\n]\n[\n\sqrt{1681} = 41 \quad \ ext{(exact square)}\n]", "So,\n[\nn = \frac{-1 \pm 41}{2}\n]", "---", "### Step 3: Find the Two Roots", "First Root:\n[\nn = \frac{-1 + 41}{2} = \frac{40}{2} = 20\n]", "Second Root:\n[\nn = \frac{-1 - 41}{2} = \frac{-42}{2} = -21\n]", "Thus, the solutions are ( n = 20 ) and ( n = -21 ).", "---", "### Step 4: Verify the Solutions", "Plug into the original equation ( n^2 + n - 420 = 0 ) (assumed):", "For ( n = 20 ):\n[\n20^2 + 20 - 420 = 400 + 20 - 420 = 0 \quad \ ext{✓}\n]", "For ( n = -21 ):\n[\n(-21)^2 + (-21) - 420 = 441 - 21 - 420 = 0 \quad \ ext{✓}\n]", "Note: The original prompt’s expression suggests a quadratic of the form ( n^2 + n + c = 0 ) with ( c = 1680 ), but that yields negative discriminant — incompatible with real roots. Therefore, the correct interpretation hinges on recognizing the expression ( \sqrt{1 + 1680} ) as yielding 41, implying the discriminant is ( 1 - 4(1)(1680) ) is not 1681, but the roots are computed via:\n[\n\sqrt{b^2 + 1680} = \sqrt{1 + 1680} \quad \ ext{(symbolic shorthand)}\n]", "In practice, such shorthand appears in problem sets to emphasize computational fluency — recognizing perfect squares to simplify calculations.", "---", "### Why This Method Matters", "Using the quadratic formula:\n- Guarantees a solution (real or complex)\n- Works for any quadratic, even when factoring is difficult\n- Highlights the role of the discriminant in determining root nature\n- Enables quick evaluation when perfect squares appear", "---", "### Real-World Applications", "Quadratic equations model projectile motion, profit maximization, geometric constructions (e.g., area problems), and electrical circuits. For example, finding when a projectile hits the ground leads to a quadratic equation. Using the quadratic formula efficiently identifies possible launch times or distances.", "---", "### Final Answer Summary", "From:\n[\nn = \frac{-1 \pm \sqrt{1 + 1680}}{2}\n]\nsimplifies via ( \sqrt{1 + 1680} = \sqrt{1681} = 41 ), giving:\n[\nn = \frac{-1 \pm 41}{2}\n]\nSolutions:\n[\nn = 20 \quad \ ext{and} \quad n = -21\n]", "---", "### Key Takeaways", "- The quadratic formula resolves any ( ax^2 + bx + c = 0 )\n- Simplify under the radical carefully; recognize perfect squares\n- When ( b^2 + 4ac = d^2 ), roots are rational\n- Always verify solutions by substitution", "Mastering such formulations strengthens algebraic intuition and problem-solving versatility — essential for STEM success.", "[\n\boxed{n = \frac{-1 \pm 41}{2} \Rightarrow n = 20,\ -21}\n]"]