Answer
$\dfrac{y+5}{2(y+1)}, \space y\ne5$.
Work Step by Step
Note that $\dfrac{y^2-25}{2y^2-8y-10}=\dfrac{y^2-5^2}{2y^2-8y-10}=\dfrac{y^2-5^2}{2(y^2-4y-5)}$
Factor $y^2-4y-5$
Rewrite $-4y$ as $-5y+1y$.
$=y^2-5y+1y-5$
Group the terms.
$=(y^2-5y)+(1y-5)$
Factor each group.
$=y(y-5)+1(y-5)$
Factor out $(y-5)$.
$=(y-5)(y+1)$
Thus, the given expression is equivalent to:
$=\dfrac{y^2-5^2}{2(y-5)(y+1)}$
Use special formula $a^2-b^2=(a+b)(a-b)$ in the numerator.
$=\dfrac{(y+5)(y-5)}{2(y-5)(y+1)}$
Cancel the common factor $y-5$ to obtain:.
$=\dfrac{y+5}{2(y+1)}, y\ne5$
Hence, the lowest term is $\dfrac{y+5}{2(y+1)}, \space y\ne5$.
