Setting \( g'(t) = 0 \) gives the equation \( 3t^2 - 8t + 4 = 0 \). Solving this quadratic equation using the quadratic formula \( t = rac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -8 \), and \( c = 4 \), we find:

["Title: Solving Quadratic Equations: Understanding the Solution of ( 3t^2 - 8t + 4 = 0 ) Using the Quadratic Formula", "When analyzing functions in calculus and algebra, determining critical points—such as where a derivative equals zero—is essential. Setting the derivative ( g'(t) = 0 ) leads to a quadratic equation that reveals key features like maxima, minima, or points of inflection. One such important equation is:", "[\n3t^2 - 8t + 4 = 0\n]", "This equation arises naturally when differentiating a cubic or higher-degree function, helping us locate where the function reaches local extrema. Solving this quadratic equation using the quadratic formula provides exact values for ( t ), unlocking insights into the function’s behavior.", "---", "### Setting the Derivative Equal to Zero", "Let ( g(t) ) be a differentiable function such that its derivative is:", "[\ng'(t) = 3t^2 - 8t + 4\n]", "To find the critical points, we solve:", "[\ng'(t) = 0 \quad \Rightarrow \quad 3t^2 - 8t + 4 = 0\n]", "---", "### Applying the Quadratic Formula", "The quadratic equation ( at^2 + bt + c = 0 ) has solutions given by:", "[\nt = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "For our equation:", "- ( a = 3 )\n- ( b = -8 )\n- ( c = 4 )", "Substitute these values into the formula:", "[\nt = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(4)}}{2(3)}\n]", "Simplify step-by-step:", "1. Compute ( -b = -(-8) = 8 )\n2. Compute discriminant:\n [\n (-8)^2 - 4(3)(4) = 64 - 48 = 16\n ]\n3. Compute square root of discriminant:\n [\n \sqrt{16} = 4\n ]\n4. Final expression:\n [\n t = \frac{8 \pm 4}{6}\n ]", "---", "### Calculating the Solutions", "Break into two cases:", "1. ( t = \frac{8 + 4}{6} = \frac{12}{6} = 2 )\n2. ( t = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3} )", "---", "### Final Answer", "The solutions to the equation ( 3t^2 - 8t + 4 = 0 ) are:", "[\n\boxed{t = 2} \quad \ ext{and} \quad \boxed{t = \frac{2}{3}}\n]", "These ( t )-values indicate where the original function ( g(t) ) has horizontal tangent lines—typically corresponding to local maxima, minima, or saddle points—making the quadratic formula an indispensable tool in optimization and calculus applications.", "---", "### Why This Matters", "Using the quadratic formula ensures precise, general solutions even when factoring is not obvious. Recognizing when and how to apply this method strengthens your ability to analyze functions, optimize results, and solve real-world problems across science, engineering, and economics.", "If you’re tackling similar critical point problems, mastering these steps empowers confident, accurate analysis.", "---", "Keywords: quadratic equation, derivative zero, ( g'(t) = 0 ), quadratic formula, local extrema, critical points, solve ( 3t^2 - 8t + 4 = 0 ), mathematical methods, calculus applications."]