#### Answer

$x=8$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
\sqrt{2x}-x+4=0
,$ use the properties of equality to isolate the radical and then square both sides. Then use the properties of equality to express the resulting equation in the form $ax^2+bx+c=0.$ Next is to factor the quadratic equation and solve for the value/s of the variable using the Zero Product Property. Finally, it is a must to do checking of the solution.
$\bf{\text{Solution Details:}}$
Using the properties of equality to isolate the radical, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\sqrt{2x}=x-4
.\end{array}
Squaring both sides and then using the special product on squaring binomials which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
2x=(x-4)^2
\\\\
2x=[(x)^2-2(x)(4)+(4)^2]
\\\\
2x=x^2-8x+16
\\\\
-x^2+(2x+8x)-16=0
\\\\
-x^2+10x-16=0
\\\\
-1(-x^2+10x-16)=(0)(-1)
\\\\
x^2-10x+16=0
.\end{array}
In the trinomial expression above, the value of $ac$ is $
1(16)=16
$ and the value of $b$ is $
-10
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-2,-8
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x^2-2x-8x+16=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^2-2x)-(8x-16)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(x-2)-8(x-2)=0
.\end{array}
Factoring the $GCF=
(x-2)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x-2)(x-8)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x-2=0
\\\\\text{OR}\\\\
x-8=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-2=0
\\\\
x=2
\\\\\text{OR}\\\\
x-8=0
\\\\
x=8
.\end{array}
Upon checking, only $
x=8
$ satisfies the original equation.