\[ x = rac{-(-5) \pm \sqrt{49}}{2 imes 2} = rac{5 \pm 7}{4} \]

["Understanding the Quadratic Formula: Solving ( x = \frac{-(-5) \pm \sqrt{49}}{2 \ imes 2} = \frac{5 \pm 7}{4} )", "When learning about quadratic equations, one of the most important tools you’ll encounter is the quadratic formula:", "[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "This powerful formula helps solve any quadratic equation of the form ( ax^2 + bx + c = 0 ). Today, we’ll explore a specific example:", "[\nx = \frac{-(-5) \pm \sqrt{49}}{2 \ imes 2} = \frac{5 \pm 7}{4}\n]", "### Breaking Down the Equation", "Let’s unpack the components of this equation step by step, starting with the context:", "The general form of a quadratic equation is:\n[\nax^2 + bx + c = 0\n]", "From the sample expression, we identify coefficients:\n- ( a = 2 ) (from ( 2 \ imes 2 = 4 ) in the denominator)\n- ( b = 5 ) (since ( -(-5) = 5 ))\n- ( c = 0 ) (implied; no constant term)", "### Step 1: Simplify the Numerator", "Plugging into the formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 0}}{2 \cdot 2}\n]", "We calculate each part:\n- ( (-5)^2 = 25 )\n- ( -4ac = -4 \cdot 2 \cdot 0 = 0 )\n- Discriminant ( D = 25 + 0 = 25 ) and ( \sqrt{25} = 7 )\n- Denominator: ( 2 \ imes 2 = 4 )", "So the equation becomes:\n[\nx = \frac{5 \pm 7}{4}\n]", "### Step 2: Solve for Two Possible Values", "Using the plus-minus notation (( \pm )), we split into two cases:", "1. First solution (using +):\n[\nx_1 = \frac{5 + 7}{4} = \frac{12}{4} = 3\n]", "2. Second solution (using –):\n[\nx_2 = \frac{5 - 7}{4} = \frac{-2}{4} = -\frac{1}{2}\n]", "### Step 3: Why This Matters", "This example illustrates how the quadratic formula efficiently finds both roots of a quadratic equation. Even when the coefficient ( c = 0 ), as here, the solution remains valid and straightforward. The ± symbol captures both solutions arising from the square root—essential for completeness.", "### Key Takeaways", "- The expression ( x = \frac{-(-5) \pm \sqrt{49}}{4} ) simplifies neatly using the quadratic formula.\n- The discriminant (( \sqrt{49} = 7 )) is positive, yielding two distinct real roots: ( x = 3 ) and ( x = -\frac{1}{2} ).\n- The equation ( 2x^2 + 5x = 0 ) factors easily as ( x(2x + 5) = 0 ), confirming roots at ( x = 0 ) and ( x = -\frac{5}{2} )—note, however, that the original example simplifies due to ( c = 0 ).", "### Conclusion", "Mastering the quadratic formula and recognizing how to simplify and interpret its components empowers you to solve a wide range of quadratic problems. From physics to engineering, mastering this skill is foundational. So next time you encounter ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), remember: it’s not just a formula—it’s a gateway to solutions!", "---", "Keywords for SEO:\nquadratic formula solution, solving quadratic equations, discriminant explained, quadratic formula step-by-step, application of quadratic formula, how to solve ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), real roots of quadratic, example solving ( \frac{5 \pm 7}{4} ).", "---", "Meta Description:\nLearn how to solve ( x = \frac{-(-5) \pm \sqrt{49}}{2 \ imes 2} = \frac{5 \pm 7}{4} ) using the quadratic formula. Step-by-step guide with full simplification and explanations."]