Find all solutions by factoring. \( 2 a^{2}-5 a=3 \)

To start, let's rearrange the equation into standard form. We need to move everything to one side: \( 2a^2 - 5a - 3 = 0 \). Next, we will factor this quadratic equation. To do that, we look for two numbers that multiply to \( (2)(-3) = -6 \) and add up to \(-5\). The numbers \(-6\) and \(1\) fit the bill, so we can rewrite the middle term: \( 2a^2 - 6a + a - 3 = 0 \). Now, we group our terms: \( (2a^2 - 6a) + (a - 3) = 0 \). Factoring each group we get: \( 2a(a - 3) + 1(a - 3) = 0 \). Now we notice that \( (a - 3) \) is a common factor: \( (2a + 1)(a - 3) = 0 \). Setting each factor equal to zero gives us our solutions: 1. \( 2a + 1 = 0 \) which simplifies to \( a = -\frac{1}{2} \). 2. \( a - 3 = 0 \) which gives \( a = 3 \). The solutions to the equation \( 2a^2 - 5a = 3 \) are \( a = -\frac{1}{2} \) and \( a = 3 \).