Since x²+6 is always positive we do not need the absolute values,
however we must restrict the right side 5x to always be positive
too.  5x > 0, therefore x > 0

    |x²+6| = 5x   x > 0
      x²+6 = 5x
   x²-5x+6 = 0
(x-2)(x-3) = 0
x-2=0; x-3=0
  x=2;   x=3

They both came out positive so we do not need to say x > 0 or to
discard a solution.  Both are solutions:

Edwin