This inequation is equivalent to the collection of two systems
`{((x^(2)-12x+30)/10gt1),(log_(2)((2x)/5)gt1):}`
`{(0lt(x^(2)-12x+30)/10lt1),(0ltlog_(2)((2x)/5)lt1):}`
On solving the first system, we have
`implies{(x^(2)-12x+20gt0),((2x)/5gt2):}`
`hArr{((x-10)(x-2)gt0),(xgt5):}`
`hArr{(xlt2 "and" xgt10),(xgt5):}`
Therefore the system has solution `x gt10`
On solving the second system, we have
`implies{(0ltx^(2)-12x+30lt10),(1lt(2x)/5lt2):}`
`hArr{(x^(2)-12x+30gt0 "and" x^(2)-12x=20lt),(5//2ltxlt5):}`
`hArr {(xlt6-sqrt(6) "and "xgt6+sqrt(6) "and" 2ltxlt10),(0lt xlt5):}`
Therefore the system has solution `2ltxlt6-sqrt(6)` combining both system, then solution of the original inequations is
` x epsilon (2,5,sqrt(6))uu(10,oo)`
