What about a set of expressions in the quartic's coefficients that discriminate between all cases?
There are 9 cases:
4 distinct real roots.
3 distinct real roots with one of them being a double root.
2 distinct double roots both real.
Triple root and and a distinct fourth root.
Quadruple root.
2 distinct real roots and two complex roots.
double real root and 2 complex roots.
2 double roots both complex.
four distinct complex roots.
Such sets are known for quadratic and cubic polynomials:
Quadratic ($ ax^2+bx+c $):
Discriminant: $b^2-4ac$
Positive for two distinct real roots, zero for double root, and negative for complex conjugate roots.
Cubic ($ ax^3+bx^2+cx+d $):
$\Delta_1=2b^3-9abc+27a^2d$ and $\Delta_2=\Delta_1^2-4(b^2-3ac)^3$.
Then:
$ \Delta_2>0 $ gives one real root and two complex roots.
$ \Delta_2<0$ gives three distinct real roots.
$ \Delta_2=0 $ but $ \Delta_1\neq 0$ gives a double root plus one different root.
$ \Delta_1=\Delta_2=0$ gives a triple root.
Are you familiar with the rational roots theorem? It says that any rational roots of a polynomial with integer coefficients must have the form $\frac{p}{q}$ where $p$ is a factor of the polynomial's constant term and $q$ is a factor of its leading term.
The constant term of your polynomial is $12$ and its leading term is $1$, so the possibilities for rational roots are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6$, and $\pm 12$. Note that these are only possibilities. You have to test them to see if they work. If you test all the possibilities, you'll find that there are four which work ($-1, 2, -2$, and $3$). Since any quartic polynomial has at most four real roots, you can be sure that you've found all of them.
Unfortunately, many quartic polynomials have roots which are not rational. For example, suppose that you can only find two roots $a$ and $b$ using the rational roots theorem. You could then divide the polynomial by the factor $x - a$ and then again by the factor $x-b$ to produce a quadratic equation (you can do this because of the remainder theorem, mentioned in another answer). Then use the quadratic formula to find the remaining roots.
So, basically, this technique is only useful when your polynomial has at least two rational roots.
There are techniques for solving any quartic equation, but they are quite advanced. You don't see them until you study field theory in depth. You might encounter the topic as an advanced undergraduate, but even then I think it's usually skipped.
Best Answer
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $x^2$ as $x\ne0$
The left hand side becomes
$$x^2+\left(\dfrac2x\right)^2+5\left(x-\dfrac2x\right)-18 =\left(x-\dfrac2x\right)^2+4+5\left(x-\dfrac2x\right)-18$$
So, we have a Quadratic Equation in $x-\dfrac2x$