Actually: 2x² -35x +30=0

["# Solving the Quadratic Equation: 2x² – 35x + 30 = 0", "When tackling algebra, quadratic equations often pose a key challenge—but understanding their solutions can unlock powerful problem-solving skills. The equation 2x² – 35x + 30 = 0 is a classic example that demonstrates how to find real roots using factoring, the quadratic formula, and verification. In this SEO-optimized article, we’ll explore how to solve precisely this quadratic, provide clear steps, and highlight its applications in real-world contexts to boost both learning and search visibility.", "---", "## Understanding Quadratic Equations", "A quadratic equation generally takes the standard form:", "[\nax² + bx + c = 0\n]", "Here, ( a ), ( b ), and ( c ) are constants, and ( a <br/>\neq 0 ). For our equation:", "- ( a = 2 )\n- ( b = -35 )\n- ( c = 30 )", "Quadratic equations can have two real roots, one real root (a repeated root), or two complex roots depending on the discriminant ( D = b² – 4ac ). Knowing this helps in choosing the best solution method.", "---", "## Method 1: Factoring the Quadratic", "Factoring is efficient when the quadratic trinomial can be written as the product of two binomials. We want:", "[\n2x² – 35x + 30 = 0\n\Rightarrow ?\n(x + m)(2x + n) = 0\n]", "We need ( m \cdot n = 2 \ imes 30 = 60 ) and ( m \cdot n ) terms summing to (-35x).", "Since ( m \cdot n = 60 ) and the middle term coefficient is (-35), try factor pairs of 60:", "Factors matching ( m + n = -35 ):", "- ( -3 ) and ( -20 ): because ( (-3) \ imes (-20) = 60 ) and ( -3 + (-20) = -23 ) → too small\n- ( -5 ) and ( -12 ): ( (-5) \ imes (-12) = 60 ), ( -5 + (-12) = -17 )\n- ( -2 ) and ( -30 ): ( (-2) \ imes (-30) = 60 ), ( -2 + (-30) = -32 )\n- ( -1 ) and ( -60 ): ( -1 + (-60) = -61 )", "But wait—what about switching signs? Since the ( x^2 ) coefficient is positive and the middle term is negative, both factors should be negative, but their sum must be negative and large in magnitude.", "Actually, let's reconsider: ( a = 2 ), so factoring as ( (2x - m)(x - n) ) might help.", "Try this form:", "[\n(2x - 5)(x - 6) = 2x² - 12x - 5x + 30 = 2x² - 17x + 30 \quad \ ext{Too low in middle term}\n]", "Try:", "[\n(2x - 2)(x - 15) = 2x² - 30x - 2x + 30 = 2x² - 32x + 30 \quad \ ext{Still off}\n]", "Try ( (2x - 3)(x - 10) = 2x² - 20x - 3x + 30 = 2x² - 23x + 30 )", "Getting closer. Try ( (2x - 5)(x - 6) = 2x² - 12x - 5x + 30 = 2x² - 17x + 30 )", "Still not matching (-35x)", "We’re looking for two numbers that multiply to (2 \ imes 30 = 60) and add to (-35). The pair:", "[\n-3 \ ext{ and } -20: (-3)(-20) = 60, -3 + (-20) = -23 \quad \ ext{too small}\n]\n[\n-5 and -12: -5×-12 = 60, -5 + -12 = -17\n]\n[\n-1 and -60: -60 + (-1) = -61\n]\n[\n-2 and -30: -2 + (-30) = -32\n]", "No pair adds to (-35). That means the quadratic doesn’t factor nicely with integers. So we turn to the quadratic formula, the universal method.", "---", "## Method 2: Using the Quadratic Formula", "For any ( ax² + bx + c = 0 ), the solutions are:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For ( 2x² – 35x + 30 = 0 ):", "- ( a = 2 )\n- ( b = -35 )\n- ( c = 30 )", "First, calculate the discriminant ( D ):", "[\nD = b^2 - 4ac = (-35)^2 - 4(2)(30) = 1225 - 240 = 985\n]", "Since ( D = 985 > 0 ), there are two distinct real roots.", "Now compute:", "[\nx = \frac{-(-35) \pm \sqrt{985}}{2(2)} = \frac{35 \pm \sqrt{985}}{4}\n]", "Thus, the two solutions are:", "[\nx_1 = \frac{35 + \sqrt{985}}{4}, \quad x_2 = \frac{35 - \sqrt{985}}{4}\n]", "---", "## Method 3: Verifying and Approximating the Roots", "Since ( \sqrt{985} ) is not a perfect square (( \sqrt{961} = 31, \sqrt{1024} = 32 )), approximate:", "[\n\sqrt{985} \approx 31.37\n]", "Now calculate:", "[\nx_1 \approx \frac{35 + 31.37}{4} = \frac{66.37}{4} \approx 16.59\n]\n[\nx_2 \approx \frac{35 - 31.37}{4} = \frac{3.63}{4} \approx 0.91\n]", "These approximate roots can help verify factoring or recheck mathematical consistency.", "---", "## Why This Equation Matters: Practical Applications", "While ( 2x² – 35x + 30 = 0 ) may appear abstract, quadratic equations model real-world scenarios:", "- Projectile motion: predicting time and height interactions.\n- Profit maximization: where ( R = -2x² + 35x - 30 ) (profit) is zero at break-even points.\n- Geometry: finding intersections between parabolas or curves.", "Understanding how to solve such equations empowers students, educators, and professionals to model and analyze such situations accurately.", "---", "## Step-by-Step Summary", "1. Identify ( a = 2 ), ( b = -35 ), ( c = 30 )\n2. Confirm ( D = b² - 4ac = 985 > 0 \Rightarrow ) two real solutions\n3. Use quadratic formula:", "[\nx = \frac{35 \pm \sqrt{985}}{4}\n]", "4. Approximate using calculator (optional)\n5. Verify by plugging back into original equation (recommended)", "---", "## Final Thoughts", "Solving 2x² – 35x + 30 = 0 strengthens algebraic skills with clear real-world applications. Whether using factoring (when possible), completing the square, or applying the quadratic formula, mastering these methods enhances problem-solving versatility. Always verify solutions, and remember: not all quadratics factor neatly—so the formula is your essential tool.", "---", "## SEO Keywords for Optimization", "Use these strategically in headings, meta description, and content:", "- solve 2x² – 35x + 30 = 0\n- quadratic equation solutions step-by-step\n- factoring quadratic with discriminant\n- real roots quadratic formula\n- quadratic formula applications\n- solve 2x² -35x +30\n- algebraic solutions: 2x² – 35x + 30 = 0\n- how to find roots of ax² + bx + c = 0\n- quadratic formulas and examples", "---", "## Further Reading", "- Understanding the Discriminant: When Roots Are Real and Distinct\n- Applying Quadratic Equations in Physics\n- Mastering the Quadratic Formula with Videos and Worksheets", "---", "By learning to solve 2x² – 35x + 30 = 0, you build a strong foundation in algebra—key to advancing in math and science. Keep practicing, and use these methods confidently across academic and real-world problems!"]