Answer
$x= 1$.
Work Step by Step
Given
\begin{equation}
\begin{aligned}
& \sqrt{x^2+3}=x+1 \\
& {[-1,6,1] \text { by }[-1,6,1]}
\end{aligned}
\end{equation}
Let
\begin{equation}
\begin{aligned}
f(x)&= \sqrt{x^2+3}\\
g(x)&=x+1\\
\end{aligned}
\end{equation}
A plot of the two function is shown in the figure. We see that the point of intersection between the graphs of the function is at $(x,y) = (1,2)$. This means that the solution is $x= 1$. Let's check.
\begin{equation}
\begin{aligned}
\sqrt{1^1+3}& \stackrel{?}{=}1+1 \\
\sqrt{4}& \stackrel{?}{=}2 \\
2&= 2\checkmark
\end{aligned}
\end{equation}
