t = rac{15 \pm \sqrt{15^2 - 4 imes 56}}{2} = rac{15 \pm \sqrt{225 - 224}}{2} = rac{15 \pm 1}{2}

["Understanding the Quadratic Solution: Solving ( t = \dfrac{15 \pm \sqrt{15^2 - 4 \ imes 56}}{2} )", "When solving quadratic equations, one of the most common tools used is the quadratic formula:", "[\nt = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "This formula becomes especially useful for equations of the form ( at^2 + bt + c = 0 ). In this article, we’ll walk through the step-by-step solution of the equation:", "[\nt = \dfrac{15 \pm \sqrt{15^2 - 4 \ imes 56}}{2}\n]", "---", "### Step 1: Identify coefficients from the formula", "The given equation matches the standard quadratic form ( at^2 + bt + c = 0 ), where:", "- ( a = 1 )\n- ( b = 15 )\n- ( c = 56 )", "---", "### Step 2: Calculate the discriminant", "The discriminant is the expression under the square root:", "[\n\Delta = b^2 - 4ac\n]", "Substituting the values:", "[\n\Delta = 15^2 - 4 \ imes 1 \ imes 56 = 225 - 224 = 1\n]", "The discriminant is positive and equals 1 — a rare and favorable result for finding real, distinct solutions.", "---", "### Step 3: Plug values into the quadratic formula", "Now substitute into the formula:", "[\nt = \dfrac{15 \pm \sqrt{1}}{2}\n]", "Since ( \sqrt{1} = 1 ), this simplifies to:", "[\nt = \dfrac{15 \pm 1}{2}\n]", "---", "### Step 4: Solve for both roots", "Break it into two cases:", "1. ( t = \dfrac{15 + 1}{2} = \dfrac{16}{2} = 8 )\n2. ( t = \dfrac{15 - 1}{2} = \dfrac{14}{2} = 7 )", "---", "### Final Answer", "The equation ( t = \dfrac{15 \pm \sqrt{15^2 - 4 \ imes 56}}{2} ) has two real solutions:", "[\nt = 7 \quad \ ext{and} \quad t = 8\n]", "---", "### Why This Equation Matters in Math and Science", "Quadratic equations like this arise in physics, engineering, economics, and data analysis. The discriminant tells us about the nature of the roots:", "- Positive discriminant (here: ( \Delta = 1 )) → two real, distinct solutions\n- Zero discriminant → one repeated real root\n- Negative discriminant → two complex conjugate roots", "Here, the positive discriminant confirms that time, distance, or cost modeled by such equations can yield two meaningful real outcomes.", "---", "### Conclusion", "By carefully analyzing coefficients, calculating the discriminant, and applying the quadratic formula, we efficiently find:", "[\nt = 7 \quad \ ext{and} \quad t = 8\n]", "This simple yet elegant algebraic solution highlights the power of the quadratic formula — a fundamental concept revisited for clarity and practical application.", "---", "Keywords: quadratic formula, discriminant, solve quadratic equation, ( t = \dfrac{15 \pm \sqrt{15^2 - 4 \ imes 56}}{2} ), real roots, algebra tutorial, discriminant analysis, math education, quadratic equations solution.", "---", "Understanding these foundational math tools strengthens problem-solving skills across STEM fields — master them to unlock more complex real-world applications!"]