The number of real roots of the equation $\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2-14x+6},\text{ is: }$

$\begin{array}{l}\sqrt{x^2-4x+3}+\sqrt{x^2-9}=\sqrt{4x^2+4x+6}\\ x^2-4x+3\ge 0\Rightarrow x\in (-\mathrm{\infty },1]\cup [3,\mathrm{\infty })\\ x^2-9\ge 0\Rightarrow x\in (-\mathrm{\infty },-3]\cup [3,\mathrm{\infty })\\ 4x^2-14x+6\ge 0\Rightarrow x\in \left(-\mathrm{\infty },\frac{1}{2}\right]\cup [3,\mathrm{\infty })\end{array}$
Domain of given equation is $x\in (-\mathrm{\infty },-3]\cup [3,\mathrm{\infty })$
Squaring the given equation we get,
$\begin{array}{l}\left(x^2-4x+3\right)+x^2-9+2\sqrt{\left(x^2-4x+3\right)\left(x^2-9\right)}=4x^2-14x+6\\ \Rightarrow \left(x^2-4x+3\right)(x+3)(x-3)=(x-3{)}^2(x-2{)}^2\end{array}$
$\Rightarrow x=3,\frac{7}{6};$

$\frac{7}{6}$is not in domain
$\therefore x=3$