Answer
The solutions are $y=5$ and $y=-3$.
Work Step by Step
$y^{2}-2y-8=7\implies y^{2}-2y-15=0$
To factor the polynomial $y^{2}+by+c$ to $(y+p)$ and $(y+q)$, we need to have $p+q=b$ and $pq=c$.
In this case, $b=-2$ and $c=-15$.
$\implies p=-5$ and $q=3$.
Then, $y^{2}-2y-15=(y-5)(y+3)$
$y^{2}-2y-15=0\implies (y-5)(y+3)=0$
Using zero-product property, we have
$y-5=0$ or $y+3=0$
$\implies y=5$ or $y=-3$
The solutions are $y=5$ and $y=-3$.
