Use the quadratic formula to solve the equation. \( 2 x^{2}+4 x+1=0 \) The solution set is \( \} \). (Simplify your answer. Tyoe an exact answer uning cadicals and ins net
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To solve the equation \( 2x^2 + 4x + 1 = 0 \) using the quadratic formula, we have: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \] Here, \( a = 2 \), \( b = 4 \), and \( c = 1 \). Plugging in these values: 1. Calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4 \cdot 2 \cdot 1 = 16 - 8 = 8 \] 2. Now substitute into the quadratic formula: \[ x = \frac{{-4 \pm \sqrt{8}}}{{2 \cdot 2}} = \frac{{-4 \pm 2\sqrt{2}}}{4} = \frac{{-2 \pm \sqrt{2}}}{2} \] Thus, the solutions are: \[ x = -1 + \frac{\sqrt{2}}{2} \quad \text{and} \quad x = -1 - \frac{\sqrt{2}}{2} \] The solution set is \(\left\{-1 + \frac{\sqrt{2}}{2}, -1 - \frac{\sqrt{2}}{2}\right\}\).
