Answer
$x=2$ and $x-1$
Work Step by Step
$\sqrt{3+x}=\sqrt{x^{2}+1}$
Let's square both sides of the equation:
$(\sqrt{3+x})^{2}=(\sqrt{x^{2}+1})^{2}$
$3+x=x^{2}+1$
Take all terms to the right side of the equation:
$x^{2}-x+1-3=0$
$x^{2}-x-2=0$
Solve by factoring:
$(x-2)(x+1)=0$
We get two solutions, which are:
$x=2$ and $x-1$
