Use the quadratic formula to solve the equation. \( 2 x^{2}-3 x=1 \) \( x=\square \) (Simplify your answer. Type an exact answer, using radicals and \( i \) as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)

To solve the equation \( 2x^2 - 3x = 1 \) using the quadratic formula, we first need to rewrite the equation in standard form, which is \( ax^2 + bx + c = 0 \). 1. **Rearranging the equation:** \[ 2x^2 - 3x - 1 = 0 \] Here, we have: - \( a = 2 \) - \( b = -3 \) - \( c = -1 \) 2. **Using the quadratic formula:** The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] 3. **Calculating the discriminant:** First, we calculate \( b^2 - 4ac \): \[ b^2 - 4ac = (-3)^2 - 4(2)(-1) = 9 + 8 = 17 \] 4. **Substituting values into the quadratic formula:** Now we substitute \( a \), \( b \), and the discriminant into the formula: \[ x = \frac{-(-3) \pm \sqrt{17}}{2(2)} = \frac{3 \pm \sqrt{17}}{4} \] 5. **Final answers:** Thus, the solutions for \( x \) are: \[ x = \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4} \] Now, I will provide the final answer in the required format. The solutions are: \[ x = \frac{3 + \sqrt{17}}{4}, \frac{3 - \sqrt{17}}{4} \]