Answer
the solution set is\[\left\{ -1\pm \sqrt{5} \right\}\].
Work Step by Step
The given equation can be written as \[{{x}^{2}}+2x-4=0\].
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=1,\,\,b=2,\text{ and }c=-1\]
Now put these values in the quadratic formula: \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
That is,
\[\begin{align}
& x=\frac{-2\pm \sqrt{{{\left( 2 \right)}^{2}}-4\times 1\times \left( -4 \right)}}{2\times 1} \\
& =\frac{-2\pm \sqrt{4+16}}{2} \\
& =\frac{-2\pm 2\sqrt{5}}{2} \\
& =-1\pm \sqrt{5}
\end{align}\]
Hence, the solution set is\[\left\{ -1\pm \sqrt{5} \right\}\].
