It's not the that much work.
1) Just calculate $x, x^2, x^4 = (x^2)^2, x^3 = x^2 * x$ to the side first. and
2) consider every number as having two "parts"; a "normal" number part and a "square root of negative one part"
$x = -5 + \sqrt{-4} = -5 + 2\sqrt{-1}$
$x^2 = 25 + 4(-1) + 20\sqrt{-1} = 21 + 20\sqrt{-1}$
$x^4 = 21^2 + 20^2(-1) + 2*20*21\sqrt{-1} = 441 - 400 + 840\sqrt{-1}= 41 + 840\sqrt{-1}$
$x^3 = (21 + 20\sqrt{-1})(-5 + 2\sqrt{-1}) = -110 + 40(-1) + (-100 + 42)\sqrt{-1} = -150 - 58\sqrt{-1}$
So $X^4 + 9x^3+35X^2-X+4 = (41 + 9(-150) + 35(21) -(-5) + 4) + \sqrt{-1}(840 + 9(-58) + 35(2) - 2) = -565 + 386\sqrt{-1}$
This is a hint to complex numbers. We call $\sqrt{-1} = i$ and all numbers are of the form $(a, b*i)$. $(a,bi) + (c,di) = (a+c, (b+d)i)$ and $(a,bi)(c,di) = (ac - bd, (ad + bc)i)$.