Answer
{$-\dfrac{3}{2}+\dfrac{\sqrt{13}}{2},-\dfrac{3}{2}-\dfrac{\sqrt{13}}{2}$}
Work Step by Step
Given: $x^2+3x-1=0$
This can be re-written as: $x^2+3x=1$
We will have to add both sides $(\dfrac{3}{2})^2$ to complete the square.
Thus, $x^2+3x+(\dfrac{3}{2})^2=1+(\dfrac{3}{2})^2$
or, $(x+\dfrac{3}{2})^2=\dfrac{13}{4}$
or, $(x+\dfrac{3}{2})=\pm\sqrt{\dfrac{13}{4}}$
Hence, our desired solution set is {$-\dfrac{3}{2}+\dfrac{\sqrt{13}}{2},-\dfrac{3}{2}-\dfrac{\sqrt{13}}{2}$}
