Use of substitutions in solving equations I am currently working on this problem, I am asked to solve the following
$x^2 - 4 - x\sqrt{x^3 + 3x} = 7$.
I am able to manipulate the above to obtain
$x^5 - x^4 +3x^3 + 22x^2 - 121 = 0$.
The problem asks us to use a substitution but I am unable to figure out what to substitute.
I tried letting $y=x^2 - 11$ and hence $y+14=x^2 + 3$, however I now have an $x^\frac{3}{2}$ which I am not sure how to work with that.
If anyone can offer a hint on where to go from there, especially a substitution that I might not be seeing, that would be extremely helpful. 

The original equation may be written $$x^2=11+x\sqrt{x^3+3x}\,.$$ For the square root to be defined means $x^3+3x\geqslant0$, which implies $x\geqslant 0$. Now, if $x\leqslant1$, then $11+x\sqrt{x^3+3x}\geqslant11>x^2$; on the other hand, if $x>1$, then $11+x\sqrt{x^3+3x}>x\sqrt{x^3}>x^2$. Either way, our equation can have no real solution.
So now we are looking for a complex solution. For convenience, let us rewrite the variable as $z$, so that the equation to which we seek a solution is $$z^2=11+z\sqrt{z^3+3z}.$$Again, in the complex case, for the square root to be meaningful, we must still have $$z^3+3z\in\Bbb R_{\geqslant0}.$$To see what this condition means, write $z=x+\mathrm iy$, where $x,y\in\Bbb R$. We have $$z^3+3z=x(x^2-3y^2+3)+\mathrm iy(3x^2-y^2+3)\in\Bbb R_{\geqslant0}.$$So, after eliminating the case when $z$ is real (i.e. $y=0$ and $z=x$), earlier shown  impossible, it follows that $3x^2-y^2+3=0$. Thus $$y=\pm\sqrt{3x^2+3}.$$ Hence $z^3+3z=-x(8x^2+6)\in\Bbb R_{\geqslant0}$, which implies that $x\leqslant0$. Now, after substituting $z=x\pm\mathrm i\sqrt{3x^2+3}$ into the base equation for $z$, and considering the real coefficient, a little simplification yields $$-2x^2=14+x\sqrt{-x(8x^2+6)}\quad \text{with}\quad x\leqslant0.$$This is impossible, since the RHS is always positive while the LHS can never be positive, which shows that our equation has no solution, real or complex.
Note: The fifth-degree polynomial equation mentioned in the question, which is obtained by squaring both sides of the equation $x^2-11=x\sqrt{x^3+3x}\,,$ introduces spurious solutions: those of the equation $$x^2-11=-x\sqrt{x^3+3x}\,,$$which has one real and two (conjugate) complex solutions. 
