the equation sqrt({3x^2+x+5})=x-3 wherte x is real, has : how many solutions ?

I have drawn the graph so that you can 'see' what Alan is telling you.

We can't insert attachments at the moment so I have placed the graph at the following address.

http://gyazo.com/7f189ccd8a3396ff1236cb171e883557

NOW I am going to look at it algebraically.

$$\sqrt{3x^2+x+5}=x-3\\\\
\mbox{square both sides}\\\\
3x^2+x+5=x^2-6x+9\\\\
2x^2+7x-4=0\\\\
2x^2+8x-1x-4=0\\\\
2x(x+4)-1(x+4)=0\\\\
(2x-1)(x+4)=0\\\\
x=1/2, x=-4\\\\$$

$$\mbox{ check x=0.5}\\\\
\sqrt{3\times 0.5^2+0.5+5}\\\\
=\sqrt{6.25}\\\\
=2.5\\\\
\ne 0.5-3 \mbox{ x=0.5 is no good}\\\\
\mbox{ check x = -4 }\\\\
\sqrt{3\times (-4)^2+-4+5}\\\\
=\sqrt{49}\\\\
=7\\\\
\ne -4-3 \mbox{ x=-4 is no good}\\\\
\mbox{ Hence the algebra also shows that there are no solutions.}\\\\
\mbox{ How about that!}$$

It has no solutions.  Draw a graph of each side of the equation against x and you will see they never intersect.