Solve $\sqrt{3+\sqrt{3+x}}=x$ 
Solve
  $$\sqrt{3+\sqrt{3+x}}=x$$

My try:
$$\sqrt{3+\sqrt{3+x}}=x \\ 
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?

We have to factorize $x^4 - 6x^2 - x + 6 = 0.$
We would observe that the values of x would be $1,-2,\frac{1±\sqrt{13}}{2}$.
