Answer
$x=2$
Work Step by Step
Squaring both sides of the given equation, $
2x=\sqrt{10+3x}
,$ then,
\begin{array}{l}\require{cancel}
\left( 2x \right)^2=\left( \sqrt{10+3x} \right)^2
\\\\
4x^2=10+3x
\\\\
4x^2-3x-10=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
4x^2-3x-10=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(-3)\pm\sqrt{(-3)^2-4(4)(-10)}}{2(4)}
\\\\=
\dfrac{3\pm\sqrt{9+160}}{8}
\\\\=
\dfrac{3\pm\sqrt{169}}{8}
\\\\=
\dfrac{3\pm\sqrt{(13)^2}}{8}
\\\\=
\dfrac{3\pm13}{8}
\\\\=
\dfrac{3-13}{8}
\text{ OR }
\dfrac{3+13}{8}
\\\\=
\dfrac{-10}{8}
\text{ OR }
\dfrac{16}{8}
\\\\=
\dfrac{-10}{8}
\text{ OR }
\dfrac{16}{8}
\\\\=
\dfrac{-5}{4}
\text{ OR }
2
.\end{array}
Upon checking, only $
x=2
$ satisfies the original equation.