Answer
the solution set is\[\left\{ -2\pm \sqrt{3} \right\}\].
Work Step by Step
The given equation can be written as \[{{x}^{2}}+4x+1=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=4,\text{ and }c=1\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-4\pm \sqrt{{{\left( 4 \right)}^{2}}-4\times 1\times \left( 1 \right)}}{2\times 1} \\
& =\frac{-4\pm \sqrt{16-4}}{2} \\
& =\frac{-4\pm \sqrt{12}}{2}
\end{align}\]
Further simplify it to get,
\[\begin{align}
& x=\frac{-4\pm \sqrt{12}}{2} \\
& =\frac{-4\pm 2\sqrt{3}}{2} \\
& =-2\pm \sqrt{3}
\end{align}\]
Hence, the solution set is\[\left\{ -2\pm \sqrt{3} \right\}\].
