Solve using the quadratic formula: \( t = rac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -4.9 \), \( b = 20 \), \( c = 50 \).

["# Solving Quadratic Equations: A Step-by-Step Guide Using the Quadratic Formula", "When solving real-world problems involving projectile motion, optimization, or motion under constant acceleration, quadratic equations often come into play. One common scenario is calculating the time ( t ) at which an object reaches a certain height. The general form of a quadratic equation used for this is:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "In this article, we’ll walk through solving a specific quadratic equation using this formula:", "[\nt = \frac{-20 \pm \sqrt{20^2 - 4(-4.9)(50)}}{2(-4.9)}\n]", "Here, the coefficients are:\n- ( a = -4.9 ) (related to gravitational acceleration, adjusted for units),\n- ( b = 20 ),\n- ( c = 50 ).", "Understanding how to apply the quadratic formula is essential for anyone studying physics, engineering, or mathematics. Let’s break down the solution step-by-step.", "---", "## Step 1: Identify Coefficients and Substitute", "The first step is to plug the coefficients into the quadratic formula. Substitute ( a = -4.9 ), ( b = 20 ), and ( c = 50 ):", "[\nt = \frac{-20 \pm \sqrt{(20)^2 - 4(-4.9)(50)}}{2(-4.9)}\n]", "This ensures we account for the correct values in each part of the formula.", "---", "## Step 2: Compute the Discriminant", "The discriminant ( D = b^2 - 4ac ) determines the nature of the solutions (real and distinct, real and repeated, or complex).", "[\nD = 20^2 - 4(-4.9)(50)\n]", "Calculate each part:\n- ( 20^2 = 400 )\n- ( -4 \ imes -4.9 \ imes 50 = 4.9 \ imes 200 = 980 )\n- So, ( D = 400 + 980 = 1380 )", "A positive discriminant indicates two real, distinct solutions — ideal for modeling tangible timing events.", "---", "## Step 3: Substitute Back into the Formula", "Now replace the discriminant in the formula:", "[\nt = \frac{-20 \pm \sqrt{1380}}{2(-4.9)} = \frac{-20 \pm \sqrt{1380}}{-9.8}\n]", "This step prepares the equation for further simplification.", "---", "## Step 4: Simplify the Square Root", "Approximate ( \sqrt{1380} ):\nSince ( 37^2 = 1369 ) and ( 38^2 = 1444 ), ( \sqrt{1380} \approx 37.15 ).", "Use this approximation for quicker calculation:", "[\nt \approx \frac{-20 \pm 37.15}{-9.8}\n]", "---", "## Step 5: Solve for Both Roots", "Compute both possibilities using ( + ) and ( - ):", "With the plus sign:\n[\nt_1 = \frac{-20 + 37.15}{-9.8} = \frac{17.15}{-9.8} \approx -1.75 \ ext{ seconds}\n]", "With the minus sign:\n[\nt_2 = \frac{-20 - 37.15}{-9.8} = \frac{-57.15}{-9.8} \approx 5.83 \ ext{ seconds}\n]", "---", "## Step 6: Interpret the Results", "Time cannot be negative, so ( t_1 \approx -1.75 ) seconds is unphysical in this context. The valid solution is:", "[\nt \approx 5.83 \ ext{ seconds}\n]", "This means the object reaches the specified height 5.83 seconds after release — crucial information for timing in motion analysis.", "---", "## Conclusion", "Using the quadratic formula ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), we solved:", "[\nt = \frac{-20 \pm \sqrt{1380}}{-9.8}\n]", "Through careful computation, we found the meaningful solution of approximately 5.83 seconds. Mastering this method allows accurate predictions in physics simulations and real-world engineering problems involving parabolic motion.", "Key takeaway: Always verify that time values are physically meaningful (non-negative) and interpret the role of the discriminant in determining real/imaginary roots.", "---\nKeywords: quadratic formula, solve quadratic equations, projectile motion, time calculation, discriminant, physics applications, real solutions, mathematical formula"]