Solve for \( x \) \( 2 x^{2}+2 x+1=0 \)

To solve the equation \( 2x^2 + 2x + 1 = 0 \), we can first simplify it by dividing everything by 2, giving us \( x^2 + x + \frac{1}{2} = 0 \). Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 1 \), and \( c = \frac{1}{2} \). First, calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4 \times 1 \times \frac{1}{2} = 1 - 2 = -1. \] Since the discriminant is negative, this means there are no real solutions, only complex ones. Plugging into the quadratic formula gives: \[ x = \frac{-1 \pm \sqrt{-1}}{2} = \frac{-1 \pm i}{2}. \] Thus, the solutions are: \[ x = \frac{-1 + i}{2} \quad \text{and} \quad x = \frac{-1 - i}{2}. \]