Answer
The solution for the equation ${{x}^{2}}+4x-1=0$ is $x=-2\pm \sqrt{5}$
Work Step by Step
On comparing ${{x}^{2}}+4x-1=0$ with the standard form of the equation,
$\begin{align}
& a=1, \\
& b=4, \\
& c=-1 \\
\end{align}$
Put these values in the formula:
$\begin{align}
& p=\frac{-4\pm \sqrt{{{4}^{2}}-4\cdot 1\cdot \left( -1 \right)}}{2\cdot 1} \\
& =\frac{-4\pm \sqrt{16-\left( -4 \right)}}{2} \\
& =\frac{-4\pm \sqrt{16+4}}{2} \\
& =\frac{-4\pm \sqrt{20}}{2}
\end{align}$
And simplifying further,
$\begin{align}
& p=\frac{-4\pm \sqrt{20}}{2} \\
& =\frac{-4}{2}\pm \frac{2\sqrt{5}}{2} \\
& =-2\pm \sqrt{5}
\end{align}$
Thus, the two solutions of the provided equations are $x=-2+\sqrt{5}$ and $x=-2-\sqrt{5}$.
