Answer
The solution set is $\left\{-\frac{2}{5}, \frac{2}{5}\right\}$.
Work Step by Step
Add $-4$ to both sides to obtain:
$25x^2-4=0 \\\\ (5x)^2-2^2=0$
RECALL:
A difference of two squares can be factored using the formula:
$a^2-b^2=(a-b)(a+b)$
Factor the difference of two squares to obtain:
$(5x-2)(5x+2)=0$
Equate each factor to 0, and then solve each equation to obtain:
$5x-2=0 \text{ or } 5x+2 = 0
\\5x=2 \text{ or } 5x=-2
\\x=\frac{2}{5} \text{ or } x=-\frac{2}{5}$
The solution set is $\left\{-\frac{2}{5}, \frac{2}{5}\right\}$.
