Answer
$x=-\dfrac{3}{2}\pm\dfrac{3\sqrt{5}}{2}$
Work Step by Step
$x-\sqrt{9-3x}=0$
Take $\sqrt{9-3x}$ to the right side:
$x=\sqrt{9-3x}$
Square both sides of the equation:
$x^{2}=(\sqrt{9-3x})^{2}$
$x^{2}=9-3x$
Take all terms to the left side of the equation:
$x^{2}+3x-9=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=1$, $b=3$ and $c=-9$.
Substitute the known values into the formula and simplify:
$x=\dfrac{-3\pm\sqrt{3^{2}-4(1)(-9)}}{2(1)}=\dfrac{-3\pm\sqrt{9+36}}{2}=...$
$...=\dfrac{-3\pm\sqrt{45}}{2}=\dfrac{-3\pm3\sqrt{5}}{2}=-\dfrac{3}{2}\pm\dfrac{3\sqrt{5}}{2}$
Neither of the solutions found makes the initial equation undefined, so the final answer is:
$x=-\dfrac{3}{2}\pm\dfrac{3\sqrt{5}}{2}$
