Answer
{$\dfrac{3}{2}-\dfrac{\sqrt{29}}{2},\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}$}
Work Step by Step
Given: $x^2-3x-5=0$
This can be re-written as: $x^2-3x=5$
We will have to add both sides $(\dfrac{-3}{2})^2$ to complete the square.
Thus, $x^2-3x+(\dfrac{-3}{2})^2=5+(\dfrac{-3}{2})^2$
or, $x^2-2(x)(\dfrac{3}{2})+\dfrac{9}{4}=\dfrac{29}{4}$
or, $(x-\dfrac{3}{2})=\pm\sqrt{\dfrac{29}{4}}$
Hence, our desired solution set is {$\dfrac{3}{2}-\dfrac{\sqrt{29}}{2},\dfrac{3}{2}+\dfrac{\sqrt{29}}{2}$}
