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.

A different viewpoint, in case you don't like the (smart) change of variables $y:=x+1$:
First a bit of calculus: Observe that $f(x):=(x+2)^4+x^4-82$ is a twice-differentiable function, with second derivative $f''(x)=12(x+2)^2+12x^2$, which is positive everywhere. Hence $f$ is convex, so it has at most two real roots. Therefore there are at most two different values such that $(x+2)^4+x^4=82$. By Descartes' sign of rules we deduce that there is exactly one positive real solution, while by the same rule applied to $f(-x)$ we find that there are either 3 or 1 negative real solutions; therefore we have two solutions, one positive and one negative.
Now a bit of playing-with-numbers: Observe that $82=81+1=3^4+1^4$. Therefore
$$(x+2)^4+x^4 = 3^4+1^4 = (1+2)^4 + 1^4$$
gives us the positive solution $x=1$. For the negative solution, due to the fact that $x^4=(-x)^4$, we can try with $x=-1$ and $x=-3$: the first one is a no-go, but the second one satisfies $x+2=-1$, so that
$$(x+2)^4+x^4 = (-1)^4 + (-3)^4 = 1^4 + 3^4 = 82.$$
Hence the solutions are $x=1$ and $x=-3$.
