Answer
$x=\frac{-3-\sqrt {13}}{2},\frac{-3+\sqrt {13}}{2}$
Work Step by Step
To complete the square, add a term that is the square of half the coefficient of $x$ to both sides of the equation, and factor the left side. Then use the square root property to simplify both sides, and then isolate $x$.
$x^2+3x-1=0$
$(\frac{3}{2})^2$
$x^2+3x-1+1+(\frac{3}{2})^2=1+(\frac{3}{2})^2$
$x^2+3x+(\frac{3}{2})^2=1+(\frac{3}{2})^2$
$x^2+3x+\frac{9}{4}=1+\frac{9}{4}$
$x^2+3x+\frac{9}{4}=\frac{13}{4}$
$(x+\frac{3}{2})^2=\frac{13}{4}$
$x+\frac{3}{2}=±\frac{\sqrt {13}}{2}$
$x=-\frac{3}{2}±\frac{\sqrt {13}}{2}$
$x=\frac{-3-\sqrt {13}}{2},\frac{-3+\sqrt {13}}{2}$
