Multiply through by 10: x3-4x2-31x+70<0.
Now look for rational zeroes of the cubic. 70=2×5×7 so 2, 5 or 7 could be a zero or their negatives.
First, let x=2: 8-16-62+70=78-78=0, so x=2 is a zero. By dividing by this zero we can reduce the cubic to a quadratic (synthetic division):
2 | 1 -4 -31 70
1 2 -4 | 70
1 -2 -35 | 0 = x2-2x-35=(x-7)(x+5).
So there are three zeroes: 2, 7, -5.
Now we have to consider the inequality for (x+5)(x-2)(x-7). The product of these three factors must be negative.
x<-5: (-)(-)(-)<0 so x<-5 is one solution;
-5<x<2: (+)(-)(-)>0 so no solution here;
2<x<7: (+)(+)(-)<0 so 2<x<7 is another solution;
x>7: (+)(+)(+)>0 so no solution here.
SOLUTION: x<-5 or 2<x<7.