#### Answer

$x=2$

#### Work Step by Step

$2x=\sqrt{10+3x}$
Square both sides of the equation:
$(2x)^{2}=(\sqrt{10+3x})^{2}$
$4x^{2}=10+3x$
Take all terms to the left side:
$4x^{2}-3x-10=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a=4$, $b=-3$ and $c=-10$.
Substitute:
$x=\dfrac{-(-3)\pm\sqrt{(-3)^{2}-4(4)(-10)}}{2(4)}=\dfrac{3\pm\sqrt{9+160}}{8}=...$
$...=\dfrac{3\pm\sqrt{169}}{8}=\dfrac{3\pm13}{8}$
The two solutions are:
$x=\dfrac{3+13}{8}=2$
$x=\dfrac{3-13}{8}=-\dfrac{5}{4}$
Finally, verify the solutions:
$2(2)=\sqrt{10+3(2)}$
$4=4$ True
$2\Big(-\dfrac{5}{4}\Big)=\sqrt{10+3\Big(-\dfrac{5}{4}\Big)}$
$-\dfrac{5}{2}=\dfrac{5}{2}$ False
The answer is $x=2$