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.

Simplify as
$$4x^2[(x+1)^2+1]=5(x+1)^2 \implies 4x^4+8x^3+3x^2-10x-5=0$$
Denoting LHS as $f(x)$, note that $f(1)=0 ~\& ~f(-1/2)=0.$ This means that by remainder theorem $$f(x) ~\text{is divisible by}~ (2x+1)(x-1)=2x^2-x-1=g(x)$$
dividing $f(x)$ by $g(x)$ we see that $$f(x)=(2x^2-x-1)(2x^2+5x+5)$$
By solving $2x^2+5x+5=0$ we get $$x=\frac{-5\pm \sqrt{-15}}{4}=\frac{-5\pm i\sqrt{15}}{4}.$$
Finally, the equation has 4 roots $$x=-1/2,1,\frac{-5\pm i\sqrt{15}}{4}.$$
