The equation $\sqrt{3x^2+x+5}=x-3$, where $x$ is real, has

$\sqrt{3x^2+x+5}=x-3 ...\left(1\right)$

We know that for a quadratic expression $a{x}^2+bx+c, a>0,$ if discriminant i.e. $D=b^2-4ac<0,$ then the quadratic is always positive for all real values of the variable.

Now, consider $3x^2+x+5$

We have, discriminant $=b^2-4ac=1-4\times 3\times 5<0$

$\therefore 3x^2+x+5>0, \forall x\in R$

Squaring $\left(1\right),$ we get,

$3x^2+x+5={\left(x-3\right)}^2$

$\Rightarrow 3x^2+x+5=x^2-6x+9$

$\Rightarrow 2x^2+7x-4=0$

$\Rightarrow \left(x+4\right)\left(2x-1\right)=0$

$\Rightarrow x=-4\mathrm{,} x=\frac{1}{2}$

But $x-3<0$ for $x=-4, \frac{1}{2}$

Hence, the equation has no solution.