How to solve this quartic equation? For the quartic equation:
$$x^4 - x^3 + 4x^2 + 3x + 5 = 0$$
I tried Ferrari so far and a few others but I just can't get its complex solutions. I know it has no real solutions. 

Let $f(x) = x^4 - x^3 +4x^2 +3x+5$. Once you know that $f(x)$ doesn't have real solutions, try some easy complex numbers like $i$, $\omega$, other $n^{th}$ roots of unity etc. Note that if we plug in $x = \omega$, where $\omega$ is the complex cube-root of unit, we get that $$f(\omega) = \omega^4 - \omega^3 +4 \omega^2 + 3 \omega + 5 = \omega - 1 + 4 \omega^2 + 3 \omega +5 = 4(\omega^2 + \omega + 1) = 0.$$
Hence, $x^2 + x + 1$ divides $f(x)$. Hence, $f(x) = (x^2 + x + 1) (x^2 + ax + b)$. This gives us that $$f(x) = x^4 + (a+1)x^3 + (1 + a + b)x^2 +(a+b)x + b$$
Hence, $a=-2$ and $b=5$. Hence, $$f(x) =  (x^2 + x + 1) (x^2 - 2x + 5)$$
Hence, the roots are $$x = \omega, \omega^2, 1 \pm 2i$$ 
