Find $x$ if $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$ 

Known that:
    $$(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$$
    What is the value of $x$



Let $a=(7+4\sqrt{3})^{x^2-8}$ then $\dfrac{1}{a}=(7-4\sqrt{3})^{x^2-8}$
So the given equation will be $a+\dfrac{1}{a}=14\implies a^2-14a+1=0$ 
Solving quadratic for $a=7+4\sqrt{3}$ or $a=7-4\sqrt{3}\implies x^2-8=1$ or $-1$ 
therefore possible values of $x=3,-3,\sqrt{7},-\sqrt{7}$
