1. Problem. Solve the quadratic equation $2x^2 - 3x - 5 = 0$. 2. Formula and rules. We use the quadratic formula. The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. Important rules: the discriminant $\Delta = b^2 - 4ac$ determines the nature of roots. If $\Delta>0$ there are two distinct real roots. If $\Delta=0$ there is one real repeated root. If $\Delta<0$ there are two complex conjugate roots. 3. Identify coefficients. For $2x^2 - 3x - 5 = 0$ we have $a=2$, $b=-3$, $c=-5$. 4. Compute the discriminant. $\Delta = b^2 - 4ac = (-3)^2 - 4\cdot 2 \cdot (-5)$. Simplify: $\Delta = 9 + 40 = 49$. 5. Apply the quadratic formula. $$x = \frac{-b \pm \sqrt{\Delta}}{2a}$$. Substitute values: $$x = \frac{-(-3) \pm \sqrt{49}}{2\cdot 2}$$. Simplify: $$x = \frac{3 \pm 7}{4}$$. 6. Evaluate the two roots. First root: $x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2}$. Second root: $x = \frac{3 - 7}{4} = \frac{-4}{4} = -1$. 7. Final answer. The solutions are $x = \frac{5}{2}$ and $x = -1$.