Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$ 
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$

I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sure how to find the root for negative values of $x$.
By multiplying both sides by $(x+1)^2$ we get,
$$x^2(x^2+2x+1)+x^2-\frac54(x^2+2x+1)=0$$
$$4x^4+8x^3+3x^2-10x-5=0 \implies (x-1)(4x^3+12x^2+15x+5)=0$$
But I can't factor the third degree polynomial.

$x^2 + \left(\frac{x}{x+1}\right)^2 = \frac{5}{4}$
Let $y=-x\qquad ⇒ y^2 + \left(\frac{y}{1-y}\right)^2 = \frac{5}{4}$
Let $z=\frac{x}{x+1}\qquad ⇒ z^2 + \left(\frac{z}{1-z}\right)^2 = \frac{5}{4}$
By inspection, $(x=1)$ is a solution.
$(x=1) → (y=-1) → (z=-1)$
Solve $z=-1$, back for x, we have $\;\left(x=-\frac{1}{2}\right)$
Original formula can be rewritten as quartic polynomial.
With 2 known roots, it reduced to simple quadratics.
