Solve for \( z \). \( 4 z^{2}-7 z-5=0 \) \( z=\frac{7+\sqrt{129}}{8} \) or \( z=\frac{7-\sqrt{129}}{8} \) \( z=\frac{-7+\sqrt{129}}{8} \) or \( z=\frac{-7-\sqrt{129}}{8} \) \( z=\frac{7+\sqrt{69}}{8} \) or \( z=\frac{7-\sqrt{69}}{8} \) \( z=\frac{-7+\sqrt{69}}{8} \) or \( z=\frac{-7-\sqrt{69}}{8} \)

To solve the quadratic equation \( 4z^2 - 7z - 5 = 0 \) using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we set \( a = 4 \), \( b = -7 \), and \( c = -5 \). Calculating the discriminant: \[ b^2 - 4ac = (-7)^2 - 4(4)(-5) = 49 + 80 = 129 \] Now substituting into the formula gives us: \[ z = \frac{7 \pm \sqrt{129}}{2 \cdot 4} = \frac{7 \pm \sqrt{129}}{8} \] Thus, the correct solutions are \( z = \frac{7 + \sqrt{129}}{8} \) or \( z = \frac{7 - \sqrt{129}}{8} \). Exploring the historical context, quadratic equations date back to ancient civilizations! The Babylonians (around 2000 BC) had methods for solving quadratic-type problems, but they didn’t necessarily use an algebraic framework. It wasn't until later that mathematicians like Al-Khwarizmi formalized and generalized these solutions, marking a significant milestone in the development of algebra. For real-world applications, solving quadratic equations is super useful! Whether you're calculating the trajectory of a soccer ball, optimizing area in landscaping, or figuring out profits in a business venture, quadratics pop up everywhere. Understanding how to solve them gives you a powerful tool for making informed decisions in various fields!