Just more algebra

Let $\large y = x^2 + x$

then

$\large y = \frac{156}{y} - 1$

Multiplying both sides by $\large y$ and rearranging the equation

$\large y^2 + y - 156 = 0$

$\large \left(y - 12\right)\left(y + 13\right) = 0$

if $\large y = 12$ then $\large x^2 + x - 12 = 0$

$\large \left(x - 3\right)\left(x + 4\right) = 0$

then $\large x = 3, -4$

if $\large y = -13$ then $\large x^2 + x + 13 = 0$ which has no real solutions

so the sum of all real solutions is $\large-4 + 3 = \boxed{-1}$