Answer
\[\left\{ -3\sqrt{3},\sqrt{3} \right\}\]
Work Step by Step
Compare the give equation with standard quadratic equation
\[\begin{align}
& a{{x}^{2}}+bx+c=0 \\
& a=1,b=2\sqrt{3},c=-9 \\
\end{align}\]
\[\begin{align}
& X=\left( \frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \right) \\
& X=\frac{-2\sqrt{3}\pm \sqrt{12-4(1)(-9)}}{2} \\
& =\frac{-2\sqrt{3\pm \sqrt{48}}}{2} \\
& =-\sqrt{3}\pm \sqrt{12}
\end{align}\]
So,
\[\begin{align}
& X=-\sqrt{3}\pm 2\sqrt{3} \\
& =-\sqrt{3}-2\sqrt{3},-\sqrt{3}+2\sqrt{3} \\
& =-3\sqrt{3},\sqrt{3}
\end{align}\]
Hence, the solution set is \[\left\{ -3\sqrt{3},\sqrt{3} \right\}\]
