I think you may have missed a term (+k?).
If you meant 8k4-74k3+64k2+k+1=0, then k=1 would be one solution, because 8-74+64+1+1=0; but there are others, one of which is real but irrational, and the other two are complex.
The given equation can be solved by Newton's iterative method. It involves calculus.
Let f(k)=8k4-74k3+64k2+1, then f'(k)=32k3-222k2+128k.
kn+1=kn-f(kn)/f'(kn) to give progressively more accurate solutions for k starting at a rough estimate for k0. We can choose k0=1 since we know that k=1 almost satisfies the equation.
k1=1-(8-74+64+1)/(32-222+128)=1-(-1)/(-62)=1-1/62=61/62.
k2=0.98338..., k3=0.9833842608, k4=0.9833842608, so one solution is 0.9834 approx.
To find more solutions we need to divide by this solution using synthetic division:
k | 8 -74 64 0 1
8 8k 8k2-74k 8k3-74k2+64k | 8k4-74k3+64k2
8 8k-74 8k2-74k+64 8k3-74k2+64k | 0 (remainder must be zero because we know k is a root).
By substituting k=0.9833842608 in 8k3+(8k-74)k2+(8k2-74k+64)k+8k3-74k2+64k=0 we now produce a cubic equation:
7.6078k3-66.1329k2-1.0341k-1.0169=0 (approximate coefficients),
and then we can differentiate and apply Newton's method as before to find another real solution (k=8.2820 approx). We then divide this real solution into the cubic so that we get a quadratic equation which can be solved using the quadratic formula to get the complex roots. Clearly this is very tedious, and it's only applicable if the question hasn't been misstated.