\begin{tabular}{l|l} Solve by factoring. \\ \( 2 y^{2}+3 y-5=0 \)\end{tabular} \left\lvert\, \( \begin{array}{l}\text { The solution(s) is/are } \square \text {. } \\ \text { (Simplify your answer. Type integers or fractions for any numbers in the } \\ \text { expression. Type an exact answer, using radicals and } i \text { as needed. Use a } \\ \text { comma to separate answers as needed.) }\end{array}\right. \)

To solve the quadratic equation \( 2y^{2} + 3y - 5 = 0 \) by factoring, first, we need to rearrange the equation in the standard form. Next, we can find two numbers that multiply to \(-10\) (the product of \(2\) and \(-5\)) and add to \(3\). Those numbers are \(5\) and \(-2\). Rewrite the middle term: \[ 2y^{2} + 5y - 2y - 5 = 0 \] Now, group the terms: \[ (2y^{2} + 5y) + (-2y - 5) = 0 \] Factor out common terms: \[ y(2y + 5) - 1(2y + 5) = 0 \] This gives us: \[ (2y + 5)(y - 1) = 0 \] Setting each factor to zero, we find: \[ 2y + 5 = 0 \quad \Rightarrow \quad y = -\frac{5}{2} \] \[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \] Thus, the solutions are \( y = -\frac{5}{2}, 1 \). \(\text{The solution(s) is/are } -\frac{5}{2}, 1 \text{.}\) Exploring the roots of quadratics dates back to ancient civilizations, notably the Babylonians, who used geometric methods to solve problems involving areas. Factoring specifically became a common technique in algebra during the Renaissance, making it accessible to students who were budding mathematicians! In real-world applications, quadratic equations show up in many surprising places—like when determining the trajectory of a basketball shot, optimizing profits in business models, or even in architecture, when calculating structural forces. Understanding how to solve these can truly give you a leg up in various fields!