Answer
$x=\left\{ \dfrac{-2-\sqrt{2}}{2},\dfrac{-2+\sqrt{2}}{2} \right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
2x^2+4x+1=0
,$ use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a=
2
, b=
4
, c=
1
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-4\pm\sqrt{4^2-4(2)(1)}}{2(2)}
\\\\
x=\dfrac{-4\pm\sqrt{16-8}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{8}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{4\cdot2}}{4}
\\\\
x=\dfrac{-4\pm\sqrt{(2)^2\cdot2}}{4}
\\\\
x=\dfrac{-4\pm2\sqrt{2}}{4}
\\\\
x=\dfrac{2(-2\pm\sqrt{2})}{4}
\\\\
x=\dfrac{\cancel2(-2\pm\sqrt{2})}{\cancel2(2)}
\\\\
x=\dfrac{-2\pm\sqrt{2}}{2}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-2-\sqrt{2}}{2}
\\\\\text{OR}\\\\
x=\dfrac{-2+\sqrt{2}}{2}
.\end{array}
Hence, $
x=\left\{ \dfrac{-2-\sqrt{2}}{2},\dfrac{-2+\sqrt{2}}{2} \right\}
.$
