t = rac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} = rac{8 \pm \sqrt{64 - 48}}{6} = rac{8 \pm \sqrt{16}}{6} = rac{8 \pm 4}{6}

["# Simplifying the Quadratic Formula: Solving ( t = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} )", "Quadratic equations are fundamental in algebra, and mastering their solution using the quadratic formula is essential for students, educators, and anyone interested in math problem-solving. In this article, we break down the calculation:", "[\nt = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}\n]", "## Step-by-Step Derivation", "Start with the standard quadratic formula:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For the equation ( at^2 + bt + c = 0 ), identify the coefficients:\n- ( a = 3 )\n- ( b = 8 )\n- ( c = 4 )", "Now substitute into the formula:", "[\nt = \frac{8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3}\n]", "Simplify each part:", "- Compute the discriminant:\n ( 8^2 = 64 )\n ( 4 \cdot 3 \cdot 4 = 48 )\n So,\n [\n \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4} = \sqrt{64 - 48} = \sqrt{16}\n ]", "- The denominator:\n ( 2 \cdot 3 = 6 )", "Putting it all together:", "[\nt = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}\n]", "## Finding the Two Roots", "Using the simplified expression:", "First root:\n[\nt_1 = \frac{8 + 4}{6} = \frac{12}{6} = 2\n]", "Second root:\n[\nt_2 = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}\n]", "Thus, the solutions are ( t = 2 ) and ( t = \frac{2}{3} ).", "## Why This Formula Matters", "The quadratic formula converts any quadratic equation into two linear solutions. This method guarantees exact or approximate roots regardless of whether the equation factors neatly. It is especially valuable when dealing with irrational or complex coefficients.", "## Key Takeaways", "- Always identify ( a ), ( b ), and ( c ) clearly.\n- Compute the discriminant ( b^2 - 4ac ) to determine solution type.\n- Simplify square roots and fractions for clear, precise answers.\n- The ± sign ensures both roots are captured.", "Understanding this process strengthens algebraic proficiency and enables confident problem-solving with quadratic equations.", "---", "TL;DR:\nThe quadratic formula ( t = \frac{8 \pm \sqrt{16}}{6} ) simplifies to two solutions: ( t = 2 ) and ( t = \frac{2}{3} ), derived by computing the discriminant and evaluating both signs of the expression."]