Absolutely! The quadratic formula is \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] Given \(a = 2\), \(b = -5\), and \(c = 3\), let's substitute these values into the quadratic formula: \[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 2 \times 3}}{2 \times 2}.\] Simplify within the square root: \[x = \frac{5 \pm \sqrt{25 - 24}}{4}.\] \[x = \frac{5 \pm \sqrt{1}}{4}.\] Now, \(\sqrt{1} = 1\), so: \[x = \frac{5 \pm 1}{4}.\] This gives two potential solutions: 1. \(x = \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}.\) 2. \(x = \frac{5 - 1}{4} = \frac{4}{4} = 1.\) Thus, the solutions to the quadratic equation \(2x^2 - 5x + 3 = 0\) are \(x = \frac{3}{2}\) and \(x = 1\).