Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -3\), \(c = -5\).

["# Solving Quadratic Equations: Step-by-Step Guide Using the Quadratic Formula", "When tackling quadratic equations, one of the most reliable tools in mathematics is the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "This formula provides exact solutions for any quadratic equation of the form ( ax^2 + bx + c = 0 ), especially when factoring is difficult or impossible. In this article, we’ll solve a specific quadratic equation using the formula and walk through each step clearly and concisely.", "---", "## The Equation: A Concrete Example", "Let’s apply the quadratic formula to the equation:", "[\n2x^2 - 3x - 5 = 0\n]", "Here, the coefficients are:\n- ( a = 2 )\n- ( b = -3 )\n- ( c = -5 )", "These integers make the calculation straightforward, but the approach works universally for any quadratics, including those with fractional or negative coefficients.", "---", "## Step 1: Identify Coefficients and Plug Into Formula", "Start by identifying ( a ), ( b ), and ( c ):\n- ( a = 2 )\n- ( b = -3 )\n- ( c = -5 )", "Now substitute these values into the quadratic formula:", "[\nx = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}\n]", "---", "## Step 2: Simplify the Expression", "Break down the expression step-by-step:", "- The numerator’s top part:\n (-(-3) = +3)", "- Inside the square root (the discriminant):\n [\n b^2 - 4ac = (-3)^2 - 4(2)(-5) = 9 + 40 = 49\n ]\n (Note: Squaring a negative gives a positive — so ((-3)^2 = 9); multiplying (4 \cdot 2 \cdot (-5) = -40), and subtracting a negative flips it to plus 40.)", "- The denominator:\n (2a = 2 \cdot 2 = 4)", "So now the equation becomes:", "[\nx = \frac{3 \pm \sqrt{49}}{4}\n]", "---", "## Step 3: Evaluate the Square Root and Final Solutions", "Since (\sqrt{49} = 7), we have:", "[\nx = \frac{3 \pm 7}{4}\n]", "This gives two possible solutions:", "1. ( x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} )\n2. ( x = \frac{3 - 7}{4} = \frac{-4}{4} = -1 )", "---", "## Step 4: Final Answer and Interpretation", "The two solutions to the equation ( 2x^2 - 3x - 5 = 0 ) are:", "[\nx = \frac{5}{2} \quad \ ext{and} \quad x = -1\n]", "These are the exact roots of the quadratic. Both values can be verified by substituting them back into the original equation to confirm they satisfy (2x^2 - 3x - 5 = 0).", "---", "## Why Use the Quadratic Formula?", "- Always correct: Unlike factoring, which fails when roots are irrational or complex, the quadratic formula guarantees real (or complex) solutions.\n- Universal applicability: Useful for any quadratic, even when coefficients are negative, fractions, or decimals.\n- Foundational: This method builds understanding for more advanced topics like solving higher-degree polynomials and analytic geometry.", "---", "## Conclusion", "Mastering the quadratic formula — (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}) — empowers students and math enthusiasts alike to confidently solve quadratic equations with confidence and precision. Whether (a), (b), and (c) are simple or complex, this reliable method delivers clear, accurate results every time.", "If you’re ready to apply this formula to your own quadratic problems, practice makes perfect — and unlock the full power of algebra!", "---", "Keywords: quadratic formula, solve quadratic equation, how to use quadratic formula, (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), step-by-step quadratic solutions, algebra tutorial, mathematical formula, quadratic equations explained", "---", "Convert numbers to words, clarify terminology like “discriminant,” and always box the final answer for emphasis on clarity and SEO value."]