Use quadratic formula: \(t = rac{4 \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2} = rac{4 \pm \sqrt{16 + 8}}{2} = rac{4 \pm \sqrt{24}}{2} = rac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}\).

["# Solve Quadratic Equations Safely: How to Use the Quadratic Formula Step-by-Step", "Mathematics is full of powerful tools, and one of the most essential for solving quadratic equations is the quadratic formula. Whether you're a student learning algebra or a professional solving complex problems, understanding how to apply this formula correctly can save time and prevent errors. In this article, we’ll explore the step-by-step use of the quadratic formula and walk through a classic example:", "[\nt = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2}\n]", "We’ll break down each part, explain the logic behind the solution, and show how to simplify expressions step by step.", "---", "## Understanding the Quadratic Formula", "The quadratic formula is a universal method for solving equations of the form:", "[\nax^2 + bx + c = 0\n]", "Where:\n- (a), (b), and (c) are coefficients\n- (x) represents the variable (often called (t) in real-world applications like time calculations)", "The formula is:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "### Deriving the Formula: A Quick Overview", "Before jumping into calculations, it helps to understand why the formula works:", "1. Start with (ax^2 + bx + c = 0)\n2. Divide by (a) to normalize: (x^2 + \frac{b}{a}x + \frac{c}{a} = 0)\n3. Rearranged and completed the square\n4. Revealed the two solution branches: (t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a})", "But for practical use, knowing how to plug values and simplify expressions efficiently is key.", "---", "## Step-by-Step Breakdown of the Given Formula", "Let’s start with the formula:", "[\nt = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2}\n]", "### Step 1: Identify coefficients", "Comparing with (ax^2 + bx + c = 0), we find:\n- (a = 1)\n- (b = 4)\n- (c = -2)", "Note: In the original equation, coefficient (a) was 1 — this explains the (-4) under the radical: (-4ac = -4(1)(-2) = +8).", "### Step 2: Compute the discriminant", "The discriminant (D) is the expression under the square root:", "[\nD = b^2 - 4ac = (-4)^2 - 4(1)(-2) = 16 + 8 = 24\n]", "A positive discriminant means two real solutions.", "### Step 3: Plug values into the formula", "[\nt = \frac{4 \pm \sqrt{24}}{2}\n]", "Now simplify (\sqrt{24}):", "[\n\sqrt{24} = \sqrt{4 \ imes 6} = 2\sqrt{6}\n]", "### Step 4: Final simplification", "Substitute:", "[\nt = \frac{4 \pm 2\sqrt{6}}{2}\n]", "Factor numerator:", "[\nt = \frac{2(2 \pm \sqrt{6})}{2} = 2 \pm \sqrt{6}\n]", "---", "## Final Answer", "[\n\boxed{t = 2 \pm \sqrt{6}}\n]", "This meaningful expression gives two solutions:\n[\nt = 2 + \sqrt{6} \quad \ ext{and} \quad t = 2 - \sqrt{6}\n]", "---", "## Why This Formula Matters", "Using the quadratic formula avoids trial-and-error methods like factoring, especially when dealing with negative or irrational coefficients. It ensures accuracy and clarity, whether you’re solving physics problems involving projectile motion, engineering equations, or optimizing real-world values.", "---", "## Key Takeaways", "- Always identify coefficients exactly\n- Calculate the discriminant to determine solution types\n- Simplify radicals step by step to avoid errors\n- Factor out common terms for clean final answers", "Mastering the quadratic formula empowers you to tackle equations with confidence — and solve them cleanly every time.", "---", "Want more? Practice applying the quadratic formula to different equations, and explore its applications in real life. If you found this guide helpful, share it with fellow learners!"]