Answer
The solution is $4$.
Work Step by Step
Subtract $2$ and add $\sqrt{x}$ to both sides of the equation to obtain:
$$x-\sqrt{x}-2+\sqrt{x}=2-2+\sqrt{x}
\\x-2=\sqrt{x}$$
Square both sides. Note that $(a-b)^2=a^2-2ab+b^2$.
$$(x-2)^2=(\sqrt{x})^2
\\x^2-2(x)(2)+2^2=x
\\x^2-4x+4=x
\\x^2-4x+4-x=0
\\x^2-5x+4=0$$
Factor the trinomial:
$$(x-4)(x-1)=0$$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&x-4=0 &\text{or} &x-1=0
\\&x=4 &\text{or} &x=1
\end{array}
Note that $1$ is does not satisfy the original equation so it is not a solution.
Therefore, the solution is $4$.
