x = rac{-(-4) \pm \sqrt{64}}{2 imes 2} = rac{4 \pm 8}{4}

["Solving Quadratic Equations Step-by-Step: Full Explanation of ( x = \frac{-(-4) \pm \sqrt{64}}{2 \ imes 2} = \frac{4 \pm 8}{4} )", "When solving quadratic equations, the quadratic formula is one of the most powerful tools in algebra. Today, we break down the solution process for the equation:", "[\nx = \frac{-(-4) \pm \sqrt{64}}{2 \ imes 2} = \frac{4 \pm 8}{4}\n]", "This form comes directly from applying the fundamental quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "In this case, comparing with the standard form ( ax^2 + bx + c = 0 ), we identify:", "- ( a = 1 ) (note: the ( 2 ) in denominator comes from expanding ( 2a ), so ( 2a = 2 \ imes 1 = 2 ))\n- ( b = 4 ) (since (-(-4) = 4))\n- ( c = 4 ) (calculated from ( \sqrt{64} = 8 ), and ( b^2 - 4ac = 16 - 16 = 0 ), though explained below)", "### Step-by-Step Breakdown: Applying the Quadratic Formula", "1. Extract coefficients\n Given:\n [\n x = \frac{-(-4) \pm \sqrt{64}}{2 \ imes 2}\n ]\n This means:\n - Coefficient of ( x^2 ): ( a = 1 )\n - Linear coefficient: ( b = 4 ) (since (-(-4) = 4))\n - Constant under the square root: ( b^2 - 4ac = 64 )", "2. Simplify numerator and denominator\n The discriminant is:\n [\n \sqrt{64} = 8\n ]\n So the equation becomes:\n [\n x = \frac{4 \pm 8}{4}\n ]", "3. Split into two solutions using ( \pm )\n The ( \pm ) symbol means we calculate two values:\n - First solution (positive radical):\n [\n x = \frac{4 + 8}{4} = \frac{12}{4} = 3\n ]\n - Second solution (negative radical):\n [\n x = \frac{4 - 8}{4} = \frac{-4}{4} = -1\n ]", "### Final Solutions", "The two solutions to the equation are:", "[\nx = 3 \quad \ ext{and} \quad x = -1\n]", "These are the roots of the quadratic equation in its simplest factored or decimal form.", "### Why This Format Matters in Algebra", "Using the quadratic formula ensures accuracy even when factoring is difficult or impossible. The given expression simplifies neatly into a clean fraction, making computation straightforward. Even though the discriminant equals zero (here, ( b^2 - 4ac = 16 - 16 = 0 )), the formula still validly provides two identical real roots: ( x = \frac{4}{4} = 1 )? Wait—actual substitution confirms:", "Actually, with:\n[\n\frac{4 \pm 8}{4}\n]\n- ( (4 + 8)/4 = 12/4 = 3 )\n- ( (4 - 8)/4 = -4/4 = -1 )", "So roots are ( x = 3 ) and ( x = -1 )", "### Pro Tip", "When solving equations like:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "always:\n- Double-check signs of ( b ) and ( c )\n- Simplify numerator and denominator carefully\n- Evaluate both ( + ) and ( - ) cases to find both roots", "### Summary", "- The expression ( x = \frac{-(-4) \pm \sqrt{64}}{2 \ imes 2} = \frac{4 \pm 8}{4} ) results from the quadratic formula\n- It correctly solves the equation with roots ( x = 3 ) and ( x = -1 )\n- Understanding this form builds strong foundation for solving any quadratic equation", "---", "Keywords: quadratic formula, solving quadratic equations, step-by-step solution, algebra examples, ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), simplified radical form, quadratic roots, equation-solving methods", "Search this article to master solving quadratic equations using the standard formula — clarity meets accuracy!"]