Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = 50 \), and \( c = -81 \):

["# Solving Quadratic Equations Using the Quadratic Formula: A Step-by-Step Example", "When it comes to solving quadratic equations, the quadratic formula is one of the most powerful and reliable tools in algebra. Whether you're a student tackling homework or a lifelong learner brushing up on math fundamentals, mastering this method is essential. In this article, we’ll explore how to solve the quadratic equation\n[\n4x^2 + 50x - 81 = 0\n]\nusing the standard quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nwith coefficients ( a = 4 ), ( b = 50 ), and ( c = -81 ). By the end, you’ll see exactly how to apply the formula step by step—no shortcuts, just clear logic.", "---", "## Step 1: Identify the Coefficients", "The first step is to identify the values of ( a ), ( b ), and ( c ) from the standard form of a quadratic equation:\n[\nax^2 + bx + c = 0\n]\nFor our equation:\n[\n4x^2 + 50x - 81 = 0\n]\nwe clearly have:\n- ( a = 4 )\n- ( b = 50 )\n- ( c = -81 )", "---", "## Step 2: Compute the Discriminant", "Before applying the quadratic formula, it’s crucial to calculate the discriminant, which determines the nature of the roots. The discriminant is given by:\n[\n\Delta = b^2 - 4ac\n]", "Substituting our values:\n[\n\Delta = (50)^2 - 4(4)(-81)\n]\n[\n\Delta = 2500 + 1296 = 3796\n]", "Since ( \Delta > 0 ), we know there are two distinct real roots. This is a key insight—your solution will involve square roots and positive real values.", "---", "## Step 3: Apply the Quadratic Formula", "Now, plug ( a ), ( b ), and ( \Delta ) into the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-50 \pm \sqrt{3796}}{2 \cdot 4}\n]\n[\nx = \frac{-50 \pm \sqrt{3796}}{8}\n]", "Before simplifying the square root, it’s smart to check if 3796 can be simplified.", "---", "## Step 4: Simplify the Square Root (Optional but Helpful)", "Let’s simplify ( \sqrt{3796} ) for a cleaner answer.", "First, factor 3796:\n3796 ÷ 4 = 949\nSo,\n[\n\sqrt{3796} = \sqrt{4 \ imes 949} = 2\sqrt{949}\n]", "Now substitute back:\n[\nx = \frac{-50 \pm 2\sqrt{949}}{8}\n]\nWe can simplify the fraction by dividing numerator and denominator by 2:\n[\nx = \frac{-25 \pm \sqrt{949}}{4}\n]", "This is the final simplified exact solution.", "---", "## Step 5: Approximate the Solutions (If Needed)", "For practical purposes, you might want decimal approximations. Using a calculator:\n[\n\sqrt{949} \approx 30.81\n]", "Then,\n[\nx_1 = \frac{-25 + 30.81}{4} = \frac{5.81}{4} \approx 1.45\n]\n[\nx_2 = \frac{-25 - 30.81}{4} = \frac{-55.81}{4} \approx -13.95\n]", "These approximate roots confirm we have two real solutions as expected.", "---", "## Why This Method Works", "The quadratic formula is derived from completing the square, offering a universal approach regardless of whether the equation factors nicely. It guarantees accurate solutions—even when dealing with irrational or complex roots—making it indispensable in algebra and advanced math.", "---", "## Summary", "Solving the equation ( 4x^2 + 50x - 81 = 0 ) using the quadratic formula involves:\n1. Identifying ( a ), ( b ), and ( c ).\n2. Calculating the discriminant to understand root nature.\n3. Applying the formula with exact or simplified radical form.\n4. Either keeping exact answers or approximating numerically.", "Final solutions:\n[\nx = \frac{-25 \pm \sqrt{949}}{4}\n]", "Mastering this formula empowers you to tackle any quadratic equation with confidence. Whether purely academic or applied in science, engineering, or finance, quadratic solutions form a foundational skill every learner should have.", "---", "Keywords: quadratic equation solution, quadratic formula, solve ( 4x^2 + 50x - 81 = 0 ), discriminant, exact solution, square root simplification, algebra tutorial, step-by-step algebra", "---", "Call to action: Practice solving other quadratics using this method—every formula is powered by consistent application!"]