Question

What is the solution set for``2x^2-3x-5=0

Answer 1

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 \( x \):

1. \( x = \frac{3 + 7}{4} = \frac{10}{4} = \frac{5}{2} \)
2. \( x = \frac{3 - 7}{4} = \frac{-4}{4} = -1 \)

Thus, the solution set for the equation \( 2x^2 - 3x - 5 = 0 \) is

\[
\boxed{\left\{ \frac{5}{2}, -1 \right\}}
\]