Answer
$\displaystyle \frac{(y-1)(y+7)}{y-5}\qquad$or$\displaystyle \quad\frac{y^{2}+6x-7}{y-5}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ equals multiplying with the reciprocal, $\displaystyle \frac{Q}{P}.$
$ \displaystyle \frac{y^{2}+4y-5}{1}\div\frac{y^{2}-25}{y+7}=\frac{y^{2}+4y-5}{1}\cdot\frac{y+7}{y^{2}-25}\qquad$... factor what you can
$\left[\begin{array}{lll}
y^{2}-25= & ..... & y^{2}+4y-5=\\
=(y-5)(y+5) & & =(y+5)(y-1)
\end{array}\right]$
$=\displaystyle \frac{(y+5)(y-1)}{1}\cdot\frac{(y+7)}{(y-5)(y+5)}\qquad$... divide out the common factors
$=\displaystyle \frac{(1)(y-1)}{1}\cdot\frac{(y+7)}{(y-5)(1)}\ $
= $\displaystyle \frac{(y-1)(y+7)}{y-5}\qquad$or$\displaystyle \quad\frac{y^{2}+6x-7}{y-5}$
