Will you cross multiply?

Let $y = \dfrac{x^{2} + 1}{x}.$ Then the given equation becomes

$y + \dfrac{1}{y} = \dfrac{29}{10} \Longrightarrow 10y^{2} - 29y + 10 = 0 \Longrightarrow (2y - 5)(5y - 2) = 0,$

and so either $y = \dfrac{5}{2}$ or $y = \dfrac{2}{5}.$

Now $y = \dfrac{x^{2} + 1}{x} = \dfrac{5}{2} \Longrightarrow 2x^{2} - 5x + 2 = 0 \Longrightarrow (2x - 1)(x - 2) = 0,$

and so either $x = 2$ or $x = \dfrac{1}{2},$ both of which satisfy the original equation.

Next, $y = \dfrac{x^{2} + 1}{x} = \dfrac{2}{5} \Longrightarrow 5x^{2} - 2x + 5 = 0,$

which has no real solutions as the discriminant $b^{2} - 4ac = (-2)^{2} - 4*5*5 = -96 \lt 0.$

Thus the sum of all real solutions $x$ is $\dfrac{1}{2} + 2 = \dfrac{5}{2} = \boxed{2.5}.$