/ 2\ log(11.0)*8*\36 - t / > 0
(8*log(11.0))*(36 - t^2) > 0
(-6.0, 6.0)
x in Interval.open(-6.00000000000000, 6.00000000000000)
And(-6.0 < t, t < 6.0)
(-6.0 < t)∧(t < 6.0)