Solve for \( x \) in the equation \( 2x^2 - 3x - 5 = 0 \).

["# Solving the Quadratic Equation: 2x² - 3x - 5 = 0", "Solving quadratic equations is a fundamental skill in algebra, widely used in science, engineering, economics, and everyday problem-solving. One of the most common equations students encounter is ( 2x^2 - 3x - 5 = 0 ). Whether you're a beginner learning how to solve quadratics or a student preparing for exams, understanding how to find the values of ( x ) is essential. This article walks you through solving ( 2x^2 - 3x - 5 = 0 ) step-by-step using the quadratic formula, factoring (when possible), and verifying solutions. Let’s dive in!", "---", "## What Is a Quadratic Equation?", "A quadratic equation is any polynomial equation of degree 2, written in the standard form:\n[\nax^2 + bx + c = 0\n]\nwhere ( a ), ( b ), and ( c ) are constants, and ( a <br/>\neq 0 ). The general solution involves finding values of ( x ) that satisfy the equation, known as the roots or solutions.", "---", "## Step 1: Identify Coefficients", "Given the equation:\n[\n2x^2 - 3x - 5 = 0\n]\nWe identify the coefficients:\n- ( a = 2 )\n- ( b = -3 )\n- ( c = -5 )", "---", "## Step 2: Choose the Best Method to Solve", "There are three primary methods for solving quadratic equations:\n1. Factoring – If the quadratic factors nicely into binomials.\n2. Completing the Square – Useful when factoring is difficult.\n3. Quadratic Formula – A reliable method for any quadratic, especially when other methods fail.", "For ( 2x^2 - 3x - 5 = 0 ), factoring is possible, so we’ll explore that first, then confirm using the quadratic formula.", "---", "## Method 1: Factoring the Quadratic", "We attempt to factor ( 2x^2 - 3x - 5 ) into the form ( (mx + n)(px + q) = 0 ).", "We look for two numbers that multiply to ( a \cdot c = 2 \cdot (-5) = -10 ) and add to ( b = -3 ).", "The numbers –5 and +2 work because:\n- ( (-5) \cdot 2 = -10 )\n- ( (-5) + 2 = -3 )", "Now rewrite the middle term using these numbers:\n[\n2x^2 - 5x + 2x - 5 = 0\n]", "Group terms:\n[\n(2x^2 - 5x) + (2x - 5) = 0\n]", "Factor each group:\n[\nx(2x - 5) + 1(2x - 5) = 0\n]", "Factor out the common binomial:\n[\n(x + 1)(2x - 5) = 0\n]", "Set each factor equal to zero:\n[\nx + 1 = 0 \quad \Rightarrow \quad x = -1\n]\n[\n2x - 5 = 0 \quad \Rightarrow \quad x = \frac{5}{2}\n]", "✅ So the solutions are ( x = -1 ) and ( x = \frac{5}{2} ).", "---", "## Method 2: Confirming with the Quadratic Formula", "When factoring is tricky, the quadratic formula guarantees a solution:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Plug in ( a = 2 ), ( b = -3 ), ( c = -5 ):\n[\nx = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}\n]\n[\nx = \frac{3 \pm \sqrt{9 + 40}}{4}\n]\n[\nx = \frac{3 \pm \sqrt{49}}{4}\n]\n[\nx = \frac{3 \pm 7}{4}\n]", "Now compute both possibilities:\n[\nx = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}\n]\n[\nx = \frac{3 - 7}{4} = \frac{-4}{4} = -1\n]", "Same solutions confirmed!", "---", "## Step 3: Verifying the Solutions", "Substitute ( x = -1 ) and ( x = \frac{5}{2} ) back into the original equation to check:", "For ( x = -1 ):\n[\n2(-1)^2 - 3(-1) - 5 = 2(1) + 3 - 5 = 2 + 3 - 5 = 0 \quad \ ext{✓}\n]", "For ( x = \frac{5}{2} ):\n[\n2\left(\frac{5}{2}\right)^2 - 3\left(\frac{5}{2}\right) - 5 = 2\left(\frac{25}{4}\right) - \frac{15}{2} - 5 = \frac{50}{4} - \frac{15}{2} - 5 = \frac{25}{2} - \frac{15}{2} - 5 = \frac{10}{2} - 5 = 5 - 5 = 0 \quad \ ext{✓}\n]", "Both values satisfy the equation.", "---", "## Step 4: Understanding the Significance of the Roots", "In real-world applications, these values may represent time, distance, cost, or physical measurements. For instance, this equation could model projectile motion or profit calculations. The two distinct real roots indicate two distinct scenarios or physical states where the relationship holds.", "---", "## Additional Insights: Graphing the Equation", "The solutions ( x = -1 ) and ( x = \frac{5}{2} ) are the ( x )-intercepts of the parabola ( y = 2x^2 - 3x - 5 ). This upward-opening parabola crosses the ( x )-axis at these two points.", "---", "## Final Answer", "The solutions to the equation ( 2x^2 - 3x - 5 = 0 ) are:\n[\n\boxed{x = -1} \quad \ ext{and} \quad \boxed{x = \frac{5}{2}}\n]", "Mastering quadratic equations empowers problem-solving across disciplines. With practice, factoring and the quadratic formula become second nature—opening doors to advanced mathematics and real-world applications.", "---", "### FAQ: Frequently Asked Questions", "Q: Can I solve 2x² - 3x - 5 = 0 by graphing?\nA: Yes! Plotting the parabola reveals where it crosses the ( x )-axis—directly giving the solutions.", "Q: Are the solutions real or complex?\nA: These are real and distinct because the discriminant ( b^2 - 4ac = 49 > 0 ).", "Q: What if the equation has no real solutions?\nA: If the discriminant is negative, solutions are complex (involving imaginary numbers). But in this case, we have two real roots.", "Q: How does this equation apply in real life?\nA: Quadratic equations model motion (e.g., height over time), optimization problems (e.g., maximizing area or profit), and circuit analysis.", "---", "Keywords: solve ( 2x^2 - 3x - 5 = 0 ), quadratic formula, factoring quadratic, solve quadratics, algebra 2, equation solutions, real roots, quadratic equations explained.", "Start solving quadratic equations today—understanding this process builds a strong foundation in mathematics!"]