Answer
$\left\{ \sqrt{7},-\sqrt{7},\sqrt{2},-\sqrt{2} \right\}$.
Work Step by Step
${{\left( {{x}^{2}}-4 \right)}^{2}}-\left( {{x}^{2}}-4 \right)-6=0$
Consider $u={{x}^{2}}-4$
${{u}^{2}}-u-6=0$
Compare the above equation with $a{{x}^{2}}+bx+c=0$
$a=1$,$b=-1$ and c = -6
$u=\frac{-b\pm \sqrt{{{b}^{^{2}}}-4ac}}{2a}$
Substitute values of a, b and c in the above equation:
$\begin{align}
& u=\frac{1\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 1 \right)\left( -6 \right)}}{2\left( 1 \right)} \\
& =\frac{1\pm \sqrt{1+24}}{2} \\
& =\frac{1\pm \sqrt{25}}{2} \\
& =\frac{1\pm 5}{2}
\end{align}$
Further,
$\begin{align}
& u=\frac{1+5}{2} \\
& =\frac{6}{2} \\
& =3
\end{align}$
$\begin{align}
& u=\frac{1-5}{2} \\
& =\frac{-4}{2} \\
& =-2
\end{align}$
Now substitute $u={{x}^{2}}-4$,
$\begin{align}
& {{x}^{2}}-4=3 \\
& {{x}^{2}}=7 \\
& x=\pm \sqrt{7}
\end{align}$
Or,
$\begin{align}
& {{x}^{2}}-4=-2 \\
& {{x}^{2}}=2 \\
& x=\pm \sqrt{2}
\end{align}$
Thus, the roots are as below,
$x=\pm \sqrt{7}$ or $x=\pm \sqrt{2}$
Therefore, the values of x are $\pm \sqrt{7}$ or $\pm \sqrt{2}$
Thus, the solution set of the equation ${{\left( {{x}^{2}}-4 \right)}^{2}}-\left( {{x}^{2}}-4 \right)-6=0$ is $\left\{ \sqrt{7},-\sqrt{7},\sqrt{2},-\sqrt{2} \right\}$
