From inspection x=5 is a solution because x^3=5x^2 and 17x=85 when x=5.
Use synthetic division to find quadratic:
5 | 1 -5 17 -85
.....1 5 0 85
.....1 0 17 | 0
Alternatively, x^3+17x-5x^2-85=x(x^2+17)-5(x^2+17)=(x-5)(x^2+17).
The other zeroes are isqrt(17) and -isqrt(17). Sqrt(17)=4.1231 approx and i is imaginary square root of -1