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I'd like to find a closed-form solution for $X,Y$ in $\mathbb{R}$ for the following system of equations:

$$\left\lbrace \begin{array}{ll} a + (X - Y) (2 + 5 X - Y) = 0 \\ b X (2 + 5 X) + Y (-2 c (1 + 3 X) + d Y) = 0\end{array} \right.$$

Using Mathematica, I can get a closed-form, however it's a few hundreds of pages long. My question is whether there is a chance of having a reasonably short expression of the roots. If not, can I get a closed-form approximation of the roots?

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    You're looking at the intersection of two curves defined by quadratics; in general, such an intersection may contain as many as 4 points, hence you're probably solving a quartic. That's generally a mess, so you probably can't expect to do a lot better unless your problem has some nice symmetries that are obvious to you but not to Mathematica. Offhand, I don't see any. – 2017-01-30
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    @JohnHughes Thank you. Would it make sense to drop the $X^2$ terms, assuming I'm interested in the intersection of the graphs for $X$ in the vicinity of $0$? – 2017-01-30
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    Moreover, as your equations depend on 4 parameters, there will be many different cases, with 4,3,2,1, 0 roots... But maybe, it could be interesting to know the motivation of you study... because for example knowing the behavior of the curve when it crosses the $y$-axis looks of very limited interest. – 2017-01-31
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    @JeanMarie My motivation comes from classical mechanics. A mechanical system, of known energy $E(X,Y)$, exhibits an instability (bifurcation) when an external force reaches a value. My goal is to have the expression of the "bifurcation point", so that I can study the influence of various parameters. To find the bifurcation point, I first use the fact that equilibrium is a minimum of energy, so a root of both $\partial_X E$ and $\partial_Y E$. The latter happens to be always $0$. The birfucation starts from $Y=0$ which gives $\partial^2_{YX}E=0$. – 2017-01-31
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    [continuation] So in the end I have two equations: $\partial_X E=0$ (equilibrium) and $\partial^2_{YY}E=0$. These are the two equations in the OP. As I said, I'm also interested in approximation (so that I can comment on the physical relevance of the bifurcation point), but I'm not sure if it makes sense just to do a Taylor series at order 1 in X (or Y) around 0 (assuming X and Y are in the neighbourhood of $0$). – 2017-01-31
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    In the last sentence of [contunation] you say "X AND Y are in the neighbourhood of 0". I think you mean "X OR Y" because, if $a \neq 0$ the fist curve doesn't pass through (0,0). – 2017-01-31
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    I think that your first action is to "drop" some parameters (making some of them constant or zero). Keeping 4 parameters is much too complicated: the behaviors can so completely differ... – 2017-01-31
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    @JeanMarie I can't really drop parameters (the mechanical system has two materials, that's four material properties, 4 dimensions, etc.). I already reduced the number of parameter by a dimensionless analysis. However, I finally found a "solution", see my own answer below. – 2017-02-02

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Such a general system of equations does not have "short" closed-form solutions, see e.g. John Hughes's comment:

You're looking at the intersection of two curves defined by quadratics; in general, such an intersection may contain as many as 4 points, hence you're probably solving a quartic. That's generally a mess, so you probably can't expect to do a lot better unless your problem has some nice symmetries that are obvious to you but not to Mathematica.

However, it might be possible to approximate the solutions in some cases. Considering ranges of values of the parameters such that the system has a solution, and that the solution corresponds to $X$ (or $Y$) sufficiently close from $0$, the second equation can be approximated by a straight line.

To illustrate my point, here (graph below) are the solution curves of the first equation (blue) and second equation (yellow) for $a =b = -1.2$, $c=d=0.3$. Both curves are hyperbolas and one of their intersection is "in the vicinity" of $X=0$ (green dot). Solving the second equation for $Y$, choosing the right solution and approximating it by a first-order Taylor expansion in $X$ in the neighbourhood of $0$ gives: $Y=bX/c$. Solving the first equation for this value of $Y$, i.e. finding the intersection between the blue hyperbola and the red straight line, gives among others the following solution:

$$X_0=\dfrac{(b - c) c + \sqrt{(c- b) c^2 ((-1 + a) b + c - 5 a c)}}{(b - 5 c) (b - c)}$$ and $Y_0=bX_0/c$. This gives the red dot below, which is reasonably close from the green dot.

Mathematica graphics

The exact intersection is at about $({0.0966, -0.330})$ [green dot] while the approximation is $(0.0864,-0.346)$ [red dot].

Of course, this can only be done for some specific values of $a,b,c,d$, but it's better than nothing.