Efficiently solving algebraic equation I would like to solve following equation:
$$15 (x+2)^{-4} = 11(x+2)^{-2} +4$$
I would first remove the negative power by adding $(x+2)^4$
Then I get 
$$15 = 11(x+2)^2 + 4(x+2)^4\\
11(x+2)^2 + 4(x+2)^4 -15 = 0$$
should I do now a quadratic equation where $d=(x+2)^2$
$$11d + 4d^2 -15 =0$$
But this would be very much work
The other way is to expand them and factorize them. but this is even more work.
Isnt there are simple way?

$$15(x+2)^{-4} = 11(x+2)^{-2} + 4$$
Multiply by $(x+2)^{4}$
$$15 = 11(x+2)^2 + 4(x+2)^4$$
Let $y = (x+2)^2$
$$15 = 11y + 4y^2$$
Subtract $15$
$$4y^2 + 11y - 15=0$$
Factor
$$(4y + 15)(y - 1)=0$$
Consider both cases:
$$(4y+15)=0$$
$$ y = -\frac{15}{4}$$
and
$$(y-1)=0$$
$$y=1$$
So, the two solutions for $y$ are $-\frac{15}{4}$ and $1$.
Substitute the values for $y$ into the expression containing $x$.
$$1 = (x+2)^2$$
$$\pm 1 = x+2$$
$$x = -2 \pm 1$$
$$x = -1, -3$$
and
$$-\frac{15}{4} = (x+2)^2$$
$$\pm \frac{\sqrt{15}i}{2} = (x+2)$$
$$x = -2 \pm \frac{\sqrt{15}i}{2}$$
So all of the solution for $x$ are:
$-1, -3, -2 + \dfrac{\sqrt{15}i}{2},$ and $-2 - \dfrac{\sqrt{15}i}{2}$
