Answer
The solution set is $\left\{-5, 5\right\}$.
Work Step by Step
Factor out the GCF (which is 2) on the left side to obtain:
$2(y^2-25)=0
\\2(y^2-5^2)=0$
Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$
to obtain:
$2(y-5)(y+5)=0$
Use the Zero-Product Property by equating each factor with a variable to zero to obtain:
$\begin{array}{ccc}
&y-5=0 &\text{ or } &y+5=0
\\&y=5 &\text{ or } &y=-5
\end{array}$
Therefore the solution set is $\left\{-5, 5\right\}$.
