t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 24}}{2} = \frac{5 \pm 1}{2}

["# Solving the Quadratic Formula: A Step-by-Step Guide to ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} )", "The quadratic formula is one of the most powerful tools in algebra, enabling you to find the solutions of any quadratic equation of the form:", "[\nat^2 + bt + c = 0\n]", "Whether you're solving equations for physics, engineering, or everyday applications, mastering this formula is essential. In this article, we’ll explore how to apply the quadratic formula step-by-step—including a breakdown of the special case illustrated by:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{25 - 24}}{2}\n]", "---", "### What is the Quadratic Formula?", "The quadratic formula gives the two roots of a quadratic equation:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "- ( a ), ( b ), and ( c ) are constants in the equation ( at^2 + bt + c = 0 )\n- The expression under the square root, ( b^2 - 4ac ), is called the discriminant\n- The ± symbol means there are two possible solutions, depending on the sign", "---", "### Step 1: Identify coefficients ( a ), ( b ), and ( c )", "For any quadratic equation, identify:", "- ( a = ) coefficient of ( t^2 )\n- ( b = ) coefficient of ( t )\n- ( c = ) constant term", "Example from the expression:", "[\nt = \frac{5 \pm \sqrt{25 - 24}}{2}\n]", "Rewriting in ( at^2 + bt + c = 0 ), we get:", "[\n1\cdot t^2 + 5t - 24 = 0\n]", "So:", "- ( a = 1 )\n- ( b = 5 )\n- ( c = -24 )", "(Note: originally, the square root was ( \sqrt{25 - 24} ), coming from ( b^2 - 4ac = 5^2 - 4(1)(-24) = 25 + 96 = 121 ), not 25.) Let’s correct that for accuracy.", "Actually:", "[\nb^2 - 4ac = (5)^2 - 4(1)(-24) = 25 + 96 = 121\n]", "Therefore:", "[\nt = \frac{-5 \pm \sqrt{121}}{2 \cdot 1} = \frac{-5 \pm 11}{2}\n]", "So the correct simplified form is:", "[\nt = \frac{-5 \pm 11}{2}\n]", "Which gives:", "- ( t = \frac{6}{2} = 3 )\n- ( t = \frac{-16}{2} = -8 )", "---", "### Step 2: Understand the Discriminant", "The discriminant ( D = b^2 - 4ac ) determines the nature of the roots:", "| Discriminant | Meaning |\n|--------------|--------------------------------|\n| ( D > 0 ) | Two distinct real roots |\n| ( D = 0 ) | One real root (a repeated root)|\n| ( D < 0 ) | Two complex roots |", "In our example, ( D = 121 > 0 ), so we expect two real, different solutions — which matches our result: ( t = 3 ) and ( t = -8 ).", "---", "### Step 3: Solving the Example: ( t = \frac{-5 \pm \sqrt{25 - 96}}{2} = \frac{-5 \pm \sqrt{-71}}{2} )? ❌", "Wait—this contradicts earlier steps! Let's clarify:", "Earlier mention of ( \sqrt{25 - 24} = \sqrt{1} ) was incorrect in linking to the discriminant with ( c = -24 ). The correct discriminant is:", "[\nD = b^2 - 4ac = 5^2 - 4(1)(-24) = 25 - (-96) = 25 + 96 = 121\n]", "Only if ( c = -24 ), the full discriminant is ( 25 - (4)(1)(-24) = 25 + 96 = 121 )", "But if ( c = -6 ), then:", "[\nD = 25 - 4(1)(-6) = 25 + 24 = 49\n]", "Let’s correct the original expression for clarity.", "Suppose we revise to:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 4(1)(-6)}}{2} = \frac{-5 \pm \sqrt{25 + 24}}{2} = \frac{-5 \pm \sqrt{49}}{2} = \frac{-5 \pm 7}{2}\n]", "Solutions:", "- ( t = \frac{2}{2} = 1 )\n- ( t = \frac{-12}{2} = -6 )", "Nonetheless, the form ( \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) remains essential for solving any quadratic.", "---", "### Why This Formula Matters", "- It works universally for any quadratic equation\n- It reveals root nature via discriminant\n- It supports quadratic applications in motion, economics, geometry, and more\n- It stands as a cornerstone of algebraic problem solving", "---", "### Practical Tips", "- Always simplify the discriminant fully\n- Watch signs of ( a ), ( b ), and ( c )\n- Remember the ± means two distinct roots, unless ( D = 0 )\n- Use the formula even when factoring isn’t obvious", "---", "### Final Summary", "The expression:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "is the quadratic formula that unlocks solutions to any quadratic equation. From the example ( t = \frac{-5 \pm \sqrt{121}}{2} ), we found two real roots via the discriminant, illustrating how this formula bridges theory and computation.", "Whether you're a student, engineer, or programmer, mastering this formula is key to solving quadratic problems efficiently and accurately.", "---", "Keywords: quadratic formula, solving quadratics, ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), discriminant, real and complex roots, algebra examples, quadratic equations explanation", "Meta description: Learn step-by-step how to use the quadratic formula ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), including solving examples like ( t = \frac{5 \pm \sqrt{25 - 4ac}}{2a} ), and explore the role of the discriminant in determining root nature."]