Solution: Factor $ 2x^2 - 7x + 3 $. Find roots: $ x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm 5}{4} \Rightarrow x = 3 $ or $ x = \frac{1}{2} $. The quadratic opens upwards, so the inequality holds between the roots: $ \boxed{\left( \frac{1}{2}, 3 \right)} $.

["Solving the Quadratic Inequality: Factor $ 2x^2 - 7x + 3 $ and Find the Solution Interval", "Factoring quadratic expressions is a fundamental skill in algebra, essential for solving equations and inequalities. One frequently encountered expression is $ 2x^2 - 7x + 3 $. Understanding how to factor this quadratic not only helps solve $ 2x^2 - 7x + 3 = 0 $ but also reveals critical information about where the expression is positive or negative.", "---", "### Step-by-Step Factoring", "To factor $ 2x^2 - 7x + 3 $, we seek two binomials of the form:\n$$\n(2x - a)(x - b)\n$$\nsuch that multiplying them gives the original trinomial. Because the leading coefficient is 2, we use a leading coefficient of 2 on one binomial to match the degree.", "We need two numbers that:\n- Multiply to $ 2 \ imes 3 = 6 $ (the product of the leading coefficient and constant term),\n- Add to $ -7 $ (the coefficient of the middle term).", "After testing factor pairs of 6, the pair $ -6 $ and $ -1 $ works:\n$$\n2x^2 - 6x - x + 3 = 0\n$$", "Grouping terms:\n$$\n(2x^2 - 6x) + (-x + 3) = 2x(x - 3) -1(x - 3) = (2x - 1)(x - 3)\n$$", "Thus,\n$$\n2x^2 - 7x + 3 = (2x - 1)(x - 3)\n$$", "---", "### Finding the Roots", "Set each factor equal to zero:\n$$\n2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2}\n$$\n$$\nx - 3 = 0 \quad \Rightarrow \quad x = 3\n$$", "The roots of the equation are $ x = \frac{1}{2} $ and $ x = 3 $. These divide the number line into intervals. Since the quadratic opens upwards (the coefficient of $ x^2 $ is positive), the expression is negative between the roots and positive outside.", "---", "### Solving the Inequality", "To determine where $ 2x^2 - 7x + 3 > 0 $, we analyze the sign of the product $ (2x - 1)(x - 3) $ across the intervals defined by the roots:", "- For $ x < \frac{1}{2} $: both factors negative → product positive\n- For $ \frac{1}{2} < x < 3 $: $ 2x - 1 > 0 $, $ x - 3 < 0 $ → product negative\n- For $ x > 3 $: both factors positive → product positive", "Thus, the inequality $ 2x^2 - 7x + 3 > 0 $ holds where the expression is positive:\n$$\nx < \frac{1}{2} \quad \ ext{or} \quad x > 3\n$$", "Since the question specifies the solution set in interval notation, the final answer is:\n$$\n\boxed{\left( -\infty, \frac{1}{2} \right) \cup \left( 3, \infty \right)}\n$$", "However, if only the roots are emphasized for inequality holds between the roots (as stated in the original problem), it's important to clarify: the expression is positive outside the interval between the roots, not inside. Yet, based on typical interpretation and the original problem’s phrasing, the relevant interval satisfying $ 2x^2 - 7x + 3 > 0 $ is as above.", "---", "### Final Thoughts", "Factoring quadratics like $ 2x^2 - 7x + 3 $ not only solves equations but reveals the behavior of the function across its domain. Whether you're solving inequalities or analyzing graphs, identifying roots and the direction of the parabola is key. Remember: upward-opening quadratics are positive outside the roots, making this a powerful tool in algebra.", "---", "Keywords: factor $ 2x^2 - 7x + 3 $, solve quadratic inequality, roots $ x = \frac{1}{2}, 3 $, quadratic opens upwards, sign of quadratic, algebra tutorial, quadratic equations."]