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

Solve the equation \( 2x^2+4x+1=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}+4x+1=0\) - step1: Solve using the quadratic formula: \(x=\frac{-4\pm \sqrt{4^{2}-4\times 2}}{2\times 2}\) - step2: Simplify the expression: \(x=\frac{-4\pm \sqrt{4^{2}-4\times 2}}{4}\) - step3: Simplify the expression: \(x=\frac{-4\pm \sqrt{8}}{4}\) - step4: Simplify the expression: \(x=\frac{-4\pm 2\sqrt{2}}{4}\) - step5: Separate into possible cases: \(\begin{align}&x=\frac{-4+2\sqrt{2}}{4}\\&x=\frac{-4-2\sqrt{2}}{4}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=\frac{-2+\sqrt{2}}{2}\\&x=\frac{-4-2\sqrt{2}}{4}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{-2+\sqrt{2}}{2}\\&x=-\frac{2+\sqrt{2}}{2}\end{align}\) - step8: Rewrite: \(x_{1}=-\frac{2+\sqrt{2}}{2},x_{2}=\frac{-2+\sqrt{2}}{2}\) The solutions to the equation \(2x^{2}+4x+1=0\) using the quadratic formula are: \[ x_{1} = -\frac{2+\sqrt{2}}{2} \] \[ x_{2} = \frac{-2+\sqrt{2}}{2} \] Therefore, the solution set is \( \{-\frac{2+\sqrt{2}}{2}, \frac{-2+\sqrt{2}}{2}\} \).