(X+2)^4+(x+4)^4=82 решите пожалуйста

Решите задачу:

$\displaystyle\tt (x+2)^4+(x+4)^4=82\\\\x+3=t\\\\(t-1)^4 + (t+1)^4 = 82\\\\t^4-4t^3+6t^2-4t+1+t^4+4t^3+6t^2+4t+1=82\\\\2t^4+12t^2+2=82\\\\2(t^4+6t^2)=80\\\\t^4+6t^2-40=0\\\\\text{D}=36+4\cdot40=36+160=196=14^2\\\\t^2=\frac{-6+14}{2}=\frac{8}2=4\\\\t^2=\frac{-6-14}2=\frac{-20}2=-10\\\\\left[\begin{array}{ccc}t^2=4\\t^2=-10\end{array}\right \rightarrow \left[\begin{array}{ccc}t=2\\t=-2\\\varnothing\end{array}\right\rightarrow\left[\begin{array}{ccc}x+3=2\\x+3=-2\end{array}\right$

$\boxed{\left[\begin{array}{ccc}x=-1\\x=-5\end{array}\right}$