#### Answer

$x=-1$ or $x=2$

#### Work Step by Step

$\sqrt{3+x}=\sqrt{x^{2}+1}$
$(\sqrt{3+x})^{2}=(\sqrt{x^{2}+1})^{2}$
$3+x=x^{2}+1$
$0=x^{2}+1-x-3$
$x^{2}-x-2=0$
$(x+1)(x-2)=0$
$x+1=0$ or $x-2=0$
$x=-1$ or $x=2$
We need to confirm the solutions.
First we confirm $x=-1$
Left side=$\sqrt{3+-1}=\sqrt{2}$
Right side=$\sqrt{(-1)^{2}+1}=\sqrt{1+1}=\sqrt{2}$
Which is true.
Next we confirm $x=2$
Left side=$\sqrt{3+2}=\sqrt{5}$
Right side=$\sqrt{2^{2}+1}=\sqrt{5}$
Which is true.