Answer
$\left\{-5, 5\right\}$
Work Step by Step
Write 25 as $5^2$ to obtain:
$x^2-5^2=0$
Factor the difference of two squares using the formula $a^2-b^2=(a-b)(a+b)$ to obtain:
$(x-5)(x+5)=0$
Use the Zero-Product Property by equating each factor to zero then solving equation to obtain:
$\begin{array}{ccc}
&x-5=0 &\text{ or } &x+5=0
\\&x=5 &\text{ or } &x=-5
\end{array}$
Thus, the solution set is $\left\{-5, 5\right\}$.
