Step 1: Look at where each part is defined.
We have $\tan^{-1}\sqrt{x(x+1)}+\sin^{-1}\sqrt{x^2+x+1}=\dfrac{\pi}{2}.$ Both square roots and the inverse functions must make sense.
Step 2: Condition from the first square root.
For $\sqrt{x(x+1)}$ to be real we need $x(x+1)\ge0.$
Step 3: Condition from the inverse sine.
$\sin^{-1}$ accepts inputs only up to $1$, so we need $\sqrt{x^2+x+1}\le1$, that is $x^2+x+1\le1.$ Since $x^2+x+1=1+x(x+1)$, this means $x(x+1)\le0.$
Step 4: Combine both conditions.
One says $x(x+1)\ge0$ and the other says $x(x+1)\le0.$ The only way both hold is $x(x+1)=0.$
Step 5: Solve and check.
$x(x+1)=0$ gives $x=0$ or $x=-1.$ Then $\sqrt{x(x+1)}=0$ and $\sqrt{x^2+x+1}=1$, so the equation reads $\tan^{-1}0+\sin^{-1}1=0+\dfrac{\pi}{2}=\dfrac{\pi}{2}.$ True for both.
Step 6: Count the solutions.
Two values work, so there are $2$ real solutions. \[ \boxed{2} \]