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

To solve the quadratic equation \( 2x^{2} + 2x + 1 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \] For our equation, \( a = 2 \), \( b = 2 \), and \( c = 1 \). Plugging these values into the formula, we get: \[ x = \frac{-2 \pm \sqrt{2^{2} - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \] Calculating the discriminant: \[ b^{2} - 4ac = 4 - 8 = -4 \] Since the discriminant is negative, we have no real solutions, but rather two complex solutions. Continuing with the calculations: \[ x = \frac{-2 \pm \sqrt{-4}}{4} = \frac{-2 \pm 2i}{4} = \frac{-1 \pm i}{2} \] Thus, the solutions are: \[ x = \frac{-1 + i}{2} \quad \text{and} \quad x = \frac{-1 - i}{2} \]