Answer
$x=16$
Work Step by Step
Isolating the radical and then squaring both sides of the given equation, $
x-2\sqrt{x}=8
,$ then,
\begin{array}{l}\require{cancel}
x-8=2\sqrt{x}
\\\\
\left( x-8 \right)^2=\left( 2\sqrt{x} \right)^2
\\\\
x^2+2(x)(-8)+(-8)^2=4x
\\\\
x^2-16x+64=4x
\\\\
x^2+(-16x-4x)+64=0
\\\\
x^2-20x+64=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
x^2-20x+64=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-20)\pm\sqrt{(-20)^2-4(1)(64)}}{2(1)}
\\\\=
\dfrac{20\pm\sqrt{400-256}}{2}
\\\\=
\dfrac{20\pm\sqrt{144}}{2}
\\\\=
\dfrac{20\pm\sqrt{(12)^2}}{2}
\\\\=
\dfrac{20\pm12}{2}
\\\\=
\dfrac{20-12}{2}
\text{ OR }
\dfrac{20+12}{2}
\\\\=
\dfrac{8}{2}
\text{ OR }
\dfrac{32}{2}
\\\\=
4
\text{ OR }
16
.\end{array}
Upon checking, only $
x=16
$ satisfies the original equation.
