Answer
$x=-\displaystyle \frac{1}{2}+\frac{\sqrt{13}}{2}$ or $x=-\displaystyle \frac{1}{2}-\frac{\sqrt{13}}{2}$
Work Step by Step
$ x^{2}+x-3=0\qquad$... use the Quadractic formula. $a=1,\ b=1,\ c=-3$
$ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad$... substitute $b$ for $1,\ a$ for $1$ and $c$ for $-3$.
$ x=\displaystyle \frac{-1\pm\sqrt{(1)^{2}-4\cdot(-3)\cdot 1}}{2\cdot 1}\qquad$... simplify.
$x=\displaystyle \frac{-1\pm\sqrt{1+12}}{2}$
$ x=\displaystyle \frac{-1\pm\sqrt{13}}{2}\qquad$... the symbol $\pm$ indicates two solutions.
$x=-\displaystyle \frac{1}{2}+\frac{\sqrt{13}}{2}$ or $x=-\displaystyle \frac{1}{2}-\frac{\sqrt{13}}{2}$
