How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.

Let $x+2=a$ and $x=b$.
Thus, $a-b=2$ and we have
$$a^4+b^4=82$$ or
$$(a^2+b^2)^2-2a^2b^2=82$$ or
$$((a-b)^2+2ab)^2-2a^2b^2=82$$ or
$$(4+2ab)^2-2a^2b^2=82$$ or $$a^2b^2+8ab-33=0,$$ which gives
$ab=-11$, $x(x+2)+11=0,$ which is impossible or
$ab=-3$, $x(x+2)=3,$ which gives the answer:
$$\{1,-3\}.$$
We can end also your way.
We need to solve
$$2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$$ or
$$x^4+4x^3+12x^2+16x-33=0$$ or
$$x^4-x^3+5x^3-5x^2+17x^2-17x+33x-33=0$$ or
$$(x-1)(x^3+5x^2+17x+33)=0$$ or
$$(x-1)(x^3+3x^2+2x^2+6x+11x+33)=0$$ or
$$(x-1)(x+3)(x^2+2x+11)=0,$$
which gives the same answer.
