\(2x^2 - 4x - 6 = 0\) では、\(a = 2\)、\(b = -4\)、\(c = -6\) です。 - United Radiology

February 23, 2026 · United Radiology

["# Solving the Quadratic Equation: (2x^2 - 4x - 6 = 0) – Step-by-Step Guide", "Solving quadratic equations is a fundamental skill in algebra, widely used in science, engineering, and mathematics. One common quadratic equation is (2x^2 - 4x - 6 = 0). In this article, we’ll explore how to solve this equation step-by-step, using standard methods, while highlighting key parameters relevant to quadratic formulas.", "---", "## Understanding the Standard Form of a Quadratic Equation", "A general quadratic equation takes the form:", "[
\nax^2 + bx + c = 0
\n]", "For the equation:
\n[
\n2x^2 - 4x - 6 = 0
\n]", "We identify the coefficients:
\n- (a = 2)
\n- (b = -4)
\n- (c = -6)", "These values are essential when applying either factoring, completing the square, or the quadratic formula, which are primary methods for solving such equations.", "---", "## Methods to Solve (2x^2 - 4x - 6 = 0)", "### Method 1: Simplification and Factoring", "First, simplify the equation by dividing every term by the leading coefficient (a = 2):", "[
\n\frac{2x^2 - 4x - 6}{2} = \frac{0}{2} \Rightarrow x^2 - 2x - 3 = 0
\n]", "Now solve the simplified equation (x^2 - 2x - 3 = 0).", "Try factoring:
\nFind two numbers that multiply to (-3) and add to (-2).
\nThose numbers are (-3) and (+1).", "So:", "[
\nx^2 - 2x - 3 = (x - 3)(x + 1) = 0
\n]", "Set each factor equal to zero:", "[
\nx - 3 = 0 \Rightarrow x = 3
\n]
\n[
\nx + 1 = 0 \Rightarrow x = -1
\n]", "Solutions: (x = 3) and (x = -1)", "---", "### Method 2: Using the Quadratic Formula", "The quadratic formula is a powerful tool for solving any quadratic equation:", "[
\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\n]", "Plug in (a = 2), (b = -4), (c = -6):", "[
\nx = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}
\n]", "Calculate each part:", "- ( -b = 4 )
\n- Discriminant:
\n [
\n (-4)^2 - 4 \cdot 2 \cdot (-6) = 16 + 48 = 64
\n ]
\n- Square root:
\n [
\n \sqrt{64} = 8
\n ]
\n- Denominator:
\n [
\n 2a = 4
\n ]", "Then:", "[
\nx = \frac{4 \pm 8}{4}
\n]", "Calculate both solutions:", "1.
\n[
\nx = \frac{4 + 8}{4} = \frac{12}{4} = 3
\n]", "2.
\n[
\nx = \frac{4 - 8}{4} = \frac{-4}{4} = -1
\n]", "Again, successive solutions: (x = 3) and (x = -1)", "---", "### Method 3: Completing the Square", "Write the equation in vertex form to identify roots:", "[
\n2x^2 - 4x - 6 = 0
\n]", "Divide by 2:", "[
\nx^2 - 2x - 3 = 0
\n]", "Move constant:", "[
\nx^2 - 2x = 3
\n]", "Take half of coefficient of (x): (-2/2 = -1), square it: ((-1)^2 = 1)", "Add 1 to both sides:", "[
\nx^2 - 2x + 1 = 3 + 1 \Rightarrow (x - 1)^2 = 4
\n]", "Take square roots:", "[
\nx - 1 = \pm 2
\n]", "Solve:", "- (x = 1 + 2 = 3)
\n- (x = 1 - 2 = -1)", "---", "## Why Understanding Coefficients (a), (b), and (c) Matters", "In quadratic equations like (2x^2 - 4x - 6 = 0), the values of (a), (b), and (c) determine:", "- The shape and direction of the parabola ((a > 0) → opens upward).
\n- The discriminant (D = b^2 - 4ac), which tells us about the number of real roots (positive = two, zero = one, negative = none).
\n- The ease of solving via factoring or formula.", "In this equation, since (D = 64 > 0), two distinct real solutions exist—confirmed by our findings.", "---", "## Final Answer", "The solutions to the quadratic equation (2x^2 - 4x - 6 = 0) are:", "[
\n\boxed{x = -1} \quad \ ext{and} \quad \boxed{x = 3}
\n]", "---", "## Pro Tips for Solving Quadratics", "- Always simplify the equation by dividing through by (a) before applying methods.
\n- Check your solution by substituting back into the original equation.
\n- Recognize when factoring is difficult and switch to the quadratic formula.
\n- Use the discriminant to predict roots’ nature without solving completely.", "---", "Understanding how to solve (2x^2 - 4x - 6 = 0) opens the door to mastering quadratics — a cornerstone skill in algebra with broad applications across science and engineering. Mastering the parameters (a), (b), (c) is key to approaching any quadratic efficiently."]

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