Question 901625
note that we do not want a square root of a negative number
we are given 
sqrt[5(5(x)^2-7)-6] =
sqrt[5(5x^2 - 7) - 6] =
sqrt[ 25x^2 -35 - 6] =
sqrt[ 25x^2 - 41]
now 25x^2 > 41
x^2 > 41/25
x > sqrt(41)/5
the interval is (sqrt(41)/5, +infinity)