## College Algebra (6th Edition)

The solutions are $x=-\sqrt{3}\pm2i$
$\sqrt{3}x^{2}+6x+7\sqrt{3}=0$ Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ In this case, $a=\sqrt{3}$, $b=6$ and $c=7\sqrt{3}$ Substitute the known values into the formula and evaluate: $x=\dfrac{-6\pm\sqrt{6^{2}-4(\sqrt{3})(7\sqrt{3})}}{2\sqrt{3}}=\dfrac{-6\pm\sqrt{36-84}}{2\sqrt{3}}=...$ $...=\dfrac{-6\pm\sqrt{-48}}{2\sqrt{3}}=\dfrac{-6\pm4\sqrt{3}i}{2\sqrt{3}}=-\dfrac{6}{2\sqrt{3}}\pm\dfrac{4\sqrt{3}}{2\sqrt{3}}i=...$ $...=-\dfrac{3}{\sqrt{3}}\pm2i=-\sqrt{3}\pm2i$