To solve the quadratic equation $2x^2 - 3x - 5 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the given equation, $a = 2$, $b = -3$, and $c = -5$. Plugging these values into the quadratic formula:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}$ $x = \frac{3 \pm \sqrt{9 + 40}}{4}$ $x = \frac{3 \pm \sqrt{49}}{4}$ $x = \frac{3 \pm 7}{4}$

This gives us two solutions:

$x = \frac{3 + 7}{4} = \frac{10}{4} = 2.5$ $x = \frac{3 - 7}{4} = \frac{-4}{4} = -1$

Therefore, the solutions to the equation $2x^2 - 3x - 5 = 0$ are $x = 2.5$ and $x = -1$.

Would you like more details or have any questions?

Here are 5 numbered questions you might consider next:

1. How do you derive the quadratic formula?
2. Can you solve another quadratic equation for me?
3. What are the applications of quadratic equations in real life?
4. How do you complete the square to solve quadratic equations?
5. What are complex roots and when do they occur in quadratic equations?

Tip: Always double-check your discriminant ($b^2 - 4ac$) to determine the nature of the roots before solving the quadratic equation.