Solving quartic equations Given the following quartic equation:
$$x^4-2x^3-7x^2+8x+12=0$$
Could anyone give some techniques required to solve any quartic equation (apart from this one) if they exist?

Notice, $$x^4-2x^3-7x^2+8x+12=0$$ substituting $x=-1$ we get $$(-1)^4-2(-1)^3-7(-1)^2+8(-1)+12=0\iff 0=0$$ Hence, $x=-1$ is a root of given quartic equation i.e. $(x+1)$ is a factor of $x^4-2x^3-7x^2+8x+12$, now we have 
$$(x+1)(x^3-3x^2-4x+12)=0$$
further factorizing $x^3-3x^2-4x+12=(x-3)(x^2-4)$
$$(x+1)(x-3)(x^2-4)=0$$ 
$$(x+1)(x-3)(x-2)(x+2)=0$$ 
Hence, the roots are  $$x=\color{red}{-2, -1, 2, 3}$$
