Step 1: Apply the quadratic formula The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For the equation \( 2x^2 - 3x - 5 = 0 \), identify:- \( a = 2 \),- \( b = -3 \),- \( c = -5 \).Step 2: Substitute coefficients into the formula \[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)}\]\[x = \frac{3 \pm \sqrt{9 + 40}}{4}\]\[x = \frac{3 \pm \sqrt{49}}{4}\]\[x = \frac{3 \pm 7}{4}\]Step 3: Calculate the two solutions for \( x \) The two solutions are:\[x_1 = \frac{3 + 7}{4} = \frac{10}{4} = 2.5\]\[x_2 = \frac{3 - 7}{4} = \frac{-4}{4} = -1\]Answer: The solutions are \( x = 2.5 \) and \( x = -1 \). This corresponds to option (3).