Answer
$y=\left\{ -5-\sqrt{51},-5+\sqrt{51} \right\}$
Work Step by Step
Using the completing the square method, the solutions of the given quadratic equation, $
y^2+10y-26=0
,$ is
\begin{array}{l}\require{cancel}
y^2+10y=26
\\\\
y^2+10y+\left( \dfrac{10}{2} \right)^2=26+\left( \dfrac{10}{2} \right)^2
\\\\
y^2+10y+25=26+25
\\\\
(y+5)^2=51
\\\\
y+5=\pm\sqrt{51}
\\\\
y=-5\pm\sqrt{51}
.\end{array}
Hence, $
y=\left\{ -5-\sqrt{51},-5+\sqrt{51} \right\}
.$
