Answer
$y=\left\{ \dfrac{3-\sqrt{29}}{2},\dfrac{3+\sqrt{29}}{2} \right\}$
Work Step by Step
In standard form, the given quadratic equations, $
y^2-3y=5
,$ is equivalent to
\begin{array}{l}\require{cancel}
y^2-3y-5=0
.\end{array}
Using the Quadratic Formula, then the solutions are
\begin{array}{l}\require{cancel}
\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
\\\\=
\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(-5)}}{2(1)}
\\\\=
\dfrac{3\pm\sqrt{9+20}}{2}
\\\\=
\dfrac{3\pm\sqrt{29}}{2}
.\end{array}
Hence, the solutions are $
y=\left\{ \dfrac{3-\sqrt{29}}{2},\dfrac{3+\sqrt{29}}{2} \right\}
.$
