5t^2 + 3t + 2 - 2t^2 - 7t - 1 = 0

["# Solving the Quadratic Equation: 5t² + 3t + 2 – (2t² + 7t + 1) = 0", "Understanding and solving quadratic equations is a fundamental skill in algebra, used widely in science, engineering, economics, and everyday problem-solving. One such quadratic expression is:", "5t² + 3t + 2 – (2t² + 7t + 1) = 0", "This article guides you step-by-step through simplifying, solving, and interpreting this equation — complete with practical examples and key insights to strengthen your algebraic foundation.", "---", "## Step 1: Simplify the Equation", "The original equation combines two quadratic expressions:", "[ 5t² + 3t + 2 - (2t² + 7t + 1) = 0 ]", "Begin by distributing the negative sign:", "[\n5t² + 3t + 2 - 2t² - 7t - 1 = 0\n]", "Now combine like terms:", "- Quadratic terms: (5t² - 2t² = 3t²)\n- Linear terms: (3t - 7t = -4t)\n- Constant terms: (2 - 1 = 1)", "Resulting in the simplified quadratic equation:", "[\n3t² - 4t + 1 = 0\n]", "---", "## Step 2: Identify Coefficients for Quadratic Formula", "The standard form of a quadratic equation is:", "[\nat² + bt + c = 0\n]", "From the simplified equation (3t² - 4t + 1 = 0), we identify:", "- (a = 3)\n- (b = -4)\n- (c = 1)", "---", "## Step 3: Apply the Quadratic Formula", "The solution to any quadratic equation is given by:", "[\nt = \frac{-b \pm \sqrt{b² - 4ac}}{2a}\n]", "Substitute the values:", "[\nt = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(1)}}{2(3)}\n]", "[\nt = \frac{4 \pm \sqrt{16 - 12}}{6}\n]", "[\nt = \frac{4 \pm \sqrt{4}}{6}\n]", "[\nt = \frac{4 \pm 2}{6}\n]", "Now compute the two solutions:", "- (t = \frac{4 + 2}{6} = \frac{6}{6} = 1)\n- (t = \frac{4 - 2}{6} = \frac{2}{6} = \frac{1}{3})", "---", "## Step 4: Verify the Solutions", "Check each solution in the original simplified equation (3t² - 4t + 1 = 0):", "- For (t = 1):\n(3(1)² - 4(1) + 1 = 3 - 4 + 1 = 0) ✅", "- For (t = \frac{1}{3}):\n(3\left(\frac{1}{3}\right)^2 - 4\left(\frac{1}{3}\right) + 1 = 3 \cdot \frac{1}{9} - \frac{4}{3} + 1 = \frac{1}{3} - \frac{4}{3} + 1 = 0) ✅", "Both solutions are verified.", "---", "## Step 5: Real-World Applications", "Equations like this frequently arise in modeling real-world situations:", "- Physics: Calculating motion trajectories where position equations involve quadratic relationships.\n- Economics: Modeling cost or revenue functions that depend non-linearly on output quantities.\n- Engineering: Analyzing structural curves or optimizing resource allocation.", "Solving such equations helps predict, control, and optimize systems effectively.", "---", "## Final Answer", "The solutions to the equation (5t² + 3t + 2 - (2t² + 7t + 1) = 0) are:", "[\n\boxed{t = 1} \quad \ ext{and} \quad \boxed{t = \frac{1}{3}}\n]", "These roots represent the values of (t) where the function (f(t) = 3t² - 4t + 1) crosses zero.", "---", "## Further Resources", "- Practice solving quadratic equations using standard and graphical methods.\n- Explore the geometric meaning of the parabola (y = 3t² - 4t + 1).\n- Apply these techniques to diverse fields such as projectile motion, area maximization, and financial modeling.", "---", "Keywords: quadratic equation, solve 3t² - 4t + 1 = 0, algebraic solutions, quadratic formula, equation simplification, t³, education algebra, real-world applications of quadratics"]