This can be simplified slightly:
160/7=(1-(1+x)-30)/x. Let f(x)=160/7-(1-(1+x)-30)/x. f(x) is indeterminate at x=0, but approaches -50/7, because:
(1-(1+x)-30)/x→(1-1+30x)/x=30 when x→0. So f(x)→160/7-30=-50/7.
The next thing to note is the large negative exponent. Also, it is even, which means that whether 1+x is positive or negative, (1+x)-30 is positive. Any quantity raised to a large negative power is small, so for the purpose of finding limits for x, treat (1+x)-30 as approximately zero:
160/7≈1/x, making x≈7/160=0.04 approximately. Also 1+x<0⇒x<-1, implying that there is a negative solution for x as well as a positive solution.
We can infer that there is a solution for 0<x<0.04. The Intermediate Value Theorem (IVT) can be used to find such a solution.
f(0.02)=0.4606; f(0.01)=-2.95; f(0.015)=-1.158, f(0.0175)=-0.328; f(0.01875)=0.0709; f(0.018125)=-0.127;...
Eventually x=0.01853 approx.
Next we need to find the lower limit for x (x<-1). We could use the same method.
f(-2)=160/7=22.857; f(-1) is indeterminate→-∞. Try the interval [-2,-1.5].
Eventually x=-1.8815 approx using IVT.
This is probably the upper limit for standard scientific calculators, although slightly better accuracy can be obtained by replacing x with x-2 and solving similarly for x. Subtracting 2 from this solution will provide the solution to the original equation. This shift moves the negative solution for x closer to the origin. Let g(x)=160/7-(1-(1-x)-30)/(x-2) then use [0,0.12] (for example) and apply IVT. If x̅ is the best value so obtained then x=x̅-2.