Answer
$x=\frac{-\sqrt{3}\pm\sqrt{15}}{2}$
$x=1.07$ or $x=-2.80$
Work Step by Step
We are given:
$x^{2}+\sqrt{3}x-3=0$
We solve using the quadratic formula ($a=1,\ b=\sqrt{3},\ c=-3$):
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$x=\frac{-\sqrt{3}\pm\sqrt{(\sqrt{3})^{2}-4*1*-3}}{2*1}$
$x=\frac{-\sqrt{3}\pm\sqrt{3+12}}{2}$
$x=\frac{-\sqrt{3}\pm\sqrt{15}}{2}$
$x=1.07$ or $x=-2.80$
