Answer
$x_{1}=\dfrac{3 + \sqrt{29}}{2}$ and $x_{2}=\dfrac{3- \sqrt{29}}{2}$
Work Step by Step
Given $y^2-3y=5 \longrightarrow y^2-3y-5=0$
$a=1, \ b=-3, \ c=-5$
Using the quadratic formula: $\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-3)\pm \sqrt{(-3)^2-4\times 1\times (-5)}}{2\times 1} = \dfrac{3\pm \sqrt{9+20}}{2} = \dfrac{3\pm \sqrt{29}}{2}$
Therefore the solutions are $x_{1}=\dfrac{3 + \sqrt{29}}{2}$ and $x_{2}=\dfrac{3- \sqrt{29}}{2}$
