Complete the square!

By completing the square, $(x-\frac{x}{x+1})^2+\frac{2x^2}{x+1}=\frac{5}{4}$ $(\frac{x^2}{x+1})^2+2 \times \frac{x^2}{x+1} = \frac{5}{4}$

Let $a = \frac{x^2}{x+1}$ , $a^2+2a = \frac{5}{4}$ $4a^2+8a-5 = 0$ $a = \frac{1}{2}, a= -\frac{5}{2}$

Substituting $a$ back into the original equation, we get $x_1=-\frac{1}{2}, x_2=1$ $x_1+x_2 = \boxed{\frac{1}{2}}$

$\begin{aligned} x^2 + \frac {x^2}{(x+1)^2} & = \frac 54 \\ x^2 + \frac {x^2-1+1}{(x+1)^2} & = \frac 54 \\ x^2 + \frac {x-1}{x+1} + \frac 1{(x+1)^2} & = \frac 54 \\ x^2 + \frac {2x}{x+1} + \frac 1{(x+1)^2} & = \frac 54 + \frac x{x+1} + \frac 1{x+1} \\ \left(x + \frac 1{x+1} \right)^2 & = \frac 94 \\ \implies x + \frac 1{x+1} & = \pm \frac 32 \\ x + 1 + \frac 1{x+1} & = \begin{cases} \dfrac 52 \\ - \dfrac 12 \end{cases} \end{aligned}$

$\implies \begin{cases} \begin{aligned} 2(x+1)^2 - 5(x+1) + 2 & = 0 \\ (2(x+1)-1)((x+1)-2) & = 0 \\ x + 1 & = \frac 12, \ 2 & \\ \implies x & = - \frac 12, \ 1 \end{aligned} \\ \begin{aligned} 2(x+1)^2 + (x+1) + 2 & = 0 & \small \red{\text{No real solution.}} \end{aligned} \end{cases}$

Therefore the sum of roots is $-\dfrac 12 + 1 = \boxed{\dfrac 12}$ .