1. The problem is to solve the quadratic equation $2x^2 - 5x + 3 = 0$. 2. We use the quadratic formula to solve equations of the form $ax^2 + bx + c = 0$: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-5$, and $c=3$. 3. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times 2 \times 3 = 25 - 24 = 1$$ 4. Since $\Delta > 0$, there are two distinct real roots. 5. Substitute values into the quadratic formula: $$x = \frac{-(-5) \pm \sqrt{1}}{2 \times 2} = \frac{5 \pm 1}{4}$$ 6. Calculate each root: - For $+$ sign: $x = \frac{5 + 1}{4} = \frac{6}{4} = 1.5$ - For $-$ sign: $x = \frac{5 - 1}{4} = \frac{4}{4} = 1$ 7. Therefore, the solutions are $x = 1.5$ and $x = 1$.