Answer
$y=\dfrac{3\pm\sqrt{29}}{2}
$
Work Step by Step
Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solution/s of the given equation, $
y^2-3y=5
,$ is/are
\begin{array}{l}\require{cancel}
y^2-3y-5=0
\\\\
y=\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(-5)}}{2(1)}
\\\\
y=\dfrac{3\pm\sqrt{9+20}}{2}
\\\\
y=\dfrac{3\pm\sqrt{29}}{2}
.\end{array}
