To solve the quadratic equation \( 2x^2 - 3x - 5 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = -3 \), and \( c = -5 \).
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot (-5) \] \[ = 9 + 40 \] \[ = 49 \]
Since the discriminant is positive, we will have two distinct real solutions. Now we can apply the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{49}}{2 \cdot 2} \] \[ = \frac{3 \pm 7}{4} \]
Now we calculate the two possible values for \( x \):
- \( x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} \)
- \( x = \frac{3 - 7}{4} = \frac{-4}{4} = -1 \)
Thus, the solution set for the equation \( 2x^2 - 3x - 5 = 0 \) is
\[ \boxed{\left{ \frac{5}{2}, -1 \right}} \]