The number of real roots of the equation ${\tan }^{-1} \left(\sqrt{\mathrm{x}(\mathrm{x}+1)}\right)+{\sin }^{-1} \left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{4}$

The given equation is ${\tan }^{-1} \left(\sqrt{\mathrm{x}(\mathrm{x}+1)}\right)+{\sin }^{-1} \left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{4}$

The possible values of $x$: 

$x^2+x\ge 0$ and $x^2+x+1\ge 0$

When $x\left(x+1\right)\ge 0$, the value of $x^2+x+1$ is more than or equal to 1

Hence, $x\left(x+1\right)=0\Rightarrow x=0,or x=-1$

When $x=0$, the left hand side of the equation is $\frac{\mathrm{\pi }}{2}$, so it is wrong

When $x=-1$, the left hand side of the equation is $\frac{\mathrm{\pi }}{2}$, so it is wrong

The number of solutions for the equation ${\tan }^{-1} \left(\sqrt{\mathrm{x}(\mathrm{x}+1)}\right)+{\sin }^{-1} \left(\sqrt{{\mathrm{x}}^2+\mathrm{x}+1}\right)=\frac{\mathrm{\pi }}{4}$ is $\boxed{\mathbf{0}}$