Answer
The $x$ intercept is $x= 2$.
Work Step by Step
Given
$$ f(x)=\sqrt{2 x-3}-\sqrt{2 x}+1$$
The x intercept of a function can be found by setting the function to zero and solving for the value(s) of $x$ that makes the function equal to zero. Hence
\begin{equation}
\begin{aligned}
f(x)&=0\\
\sqrt{2 x-3}-\sqrt{2 x}+1&=0\\\
\sqrt{2 x-3}+1& =\sqrt{2 x} \\
\left(\sqrt{2 x-3}+1\right)^2& =\left(\sqrt{2 x}\right)^2\\
2 x-3+2\sqrt{2 x-3}+1& =2x\\
2\sqrt{2 x-3}& =2x-2 x+2\\
2\sqrt{2 x-3}& =2\\
\sqrt{2 x-3}& =1\\
\left(\sqrt{2 x-3}\right)^2& =1\\
2x-3&= 1\\
2x&= 1+3\\
2x&= 4\\
x&= 2
\end{aligned}
\end{equation}
Check:
In line 3, I have used the fact that $(a-b)^2= a^2-2ab+b^2$ to expand $\left(\sqrt{2 x-3}+1\right)^2= 2 x-3+2\sqrt{2 x-3}+1$.
\begin{equation}
\begin{aligned}
f(2)&= \sqrt{2 \cdot 2-3}-\sqrt{2 \cdot 2}+1\\
&= \sqrt{4-3}-\sqrt{4}+1\\
&=1-2+1\\
&= 0
\end{aligned}
\end{equation}
