This cubic has only one real root. A graph reveals the root is near to x=30.
Newton's Method can be used to find the root using calculus.
f(x)=x3-70x2+1601.375x-12285.5; f'(x)=3x2-140x+1601.375.
xn+1=xn-f(xn)/f'(xn) where x0=30.
x1=32.4093..., x2=31.8047..., x3=31.7516..., x4=31.7512..., x5=31.75129779, x6=31.75129779, so we take this as a fairly accurate value for the real root.
To find the complex roots we need the quadratic, which can be found using synthetic division by the real root:
31.75129779 | 1 -70 1601.375 -12285.5
1 31.75129779 -1214.445634 | 12285.5
1 -38.24870221 386.929066 | 0 =
x2-38.24870221x+386.929066
We can now find the complex roots:
x=(38.24870221±i√84.75304352)/2=19.12435±4.60307i approx.