A quadratic equation is given as \(ax^2 + bx + c = 0\). If \(a = 2\), \(b = -4\), and \(c = -6\), find the roots of the equation using the quadratic formula.
["## Solving a Quadratic Equation: Finding Roots with the Quadratic Formula", "A quadratic equation is a fundamental concept in algebra, defined by the standard form:\n[ ax^2 + bx + c = 0 ]\nSolving for (x) is essential in many real-world applications—from physics to engineering. One of the most powerful tools for finding the roots of a quadratic equation is the quadratic formula:\n[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]", "### Given Values", "In this example, we are given:\n- ( a = 2 )\n- ( b = -4 )\n- ( c = -6 )", "Our goal is to plug these values into the quadratic formula and solve for (x).", "### Step 1: Plug in the coefficients", "Start by substituting (a = 2), (b = -4), and (c = -6) into the quadratic formula:\n[\nx = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}\n]", "Simplify the terms step by step.", "### Step 2: Simplify the numerator", "First, compute (-(-4)):\n[\n-(-4) = 4\n]", "Next, calculate the discriminant (b^2 - 4ac):\n[\n(-4)^2 = 16\n]\n[\n4ac = 4(2)(-6) = -48\n]\n[\nb^2 - 4ac = 16 - (-48) = 16 + 48 = 64\n]", "Now the expression becomes:\n[\nx = \frac{4 \pm \sqrt{64}}{4}\n]", "### Step 3: Simplify the square root", "[\n\sqrt{64} = 8\n]", "So now:\n[\nx = \frac{4 \pm 8}{4}\n]", "### Step 4: Solve for both roots", "Use the "(\pm)" to find the two solutions:", "First root (using the plus sign):\n[\nx = \frac{4 + 8}{4} = \frac{12}{4} = 3\n]", "Second root (using the minus sign):\n[\nx = \frac{4 - 8}{4} = \frac{-4}{4} = -1\n]", "### Final Answer", "The roots of the quadratic equation (2x^2 - 4x - 6 = 0) are:\n[\nx = 3 \quad \ ext{and} \quad x = -1\n]", "These solutions can be verified by substituting (x = 3) and (x = -1) back into the original equation, confirming both satisfy it. Using the quadratic formula efficiently finds all real roots cleanly and accurately, especially when factoring is difficult or impossible.", "Whether you're an educator, student, or math enthusiast, mastering this method is key to tackling quadratic problems with confidence."]
