I think you mean 55b+7.5b1.6=300.
It can be slightly simplified: 11b+1.5b1.6=60.
I suspect that there are two solutions for b.
Let f(b)=11b+1.5b1.6-60, so df/db=11+2.4b0.6.
Newton's Method can be used to find solutions:
bn+1=bn-f(bn)/f'(bn), which is an iterative process. Let b0=0, then:
b1=-60/11, b2=16.88..., b3=5.93..., b4=4.20..., b5=4.1338..., b6=4.1337..., b7=4.13372234..., b8=4.1337223471.
A graph of the function appears to show another solution around b=-35, so let b0=-35:
b1=-35.194..., b2=-35.1936..., b3=-35.19362784..., b4=-35.193627838.
So to 4 decimal places b=4.1337 or -35.1936. When b is negative, b1.6=b8/5=5√b8=298.0866; 7.5b1.6=2235.6495; 55b=-1935.6495, therefore 55b+7.5b1.6=300, so it is a valid solution.