In this case make the substitution $y=x+4$, which transforms the equation into
$$(y-1)^4+(y+1)^4=20.$$
This is perhaps the simplest form of the equation under a linear transformation as the two factors are now at least 'similar'.

If I expand these brackets then the coefficients will match but with some opposing signs, so much cancellation will occur:
$$(y^4-4y^3+6y^2-y+1)+(y^4+4y^3+6y^2+y+1)=20.$$
So, the odd factors cancel to give
$$2y^4+12y^2-18=0$$
This then becomes the simple equation
$$y^4+6y^2-9=0\,,$$
which is a quadratic equation in $y^2$ with solutions
$$y^2 = \frac{-6\pm\sqrt{36+36}}{2}=-3(1\pm\sqrt 2).$$
One of these solutions is positive; taking the square root gives two real values for $y$ as
$$y = \pm \sqrt{3\left(\sqrt{2}-1)\right)}$$
Therefore two real solutions to the equation are
$$x= -4 \pm \sqrt{3\left(\sqrt{2}-1\right)}$$