Answer
The solution set is $\left\{-\frac{1}{3}, \frac{1}{3}\right\}$.
Work Step by Step
Add $45x^2-5$ to both sides to find:
$0=45x^2-5$
Divide by $5$ on both sides to find:
$0=9x^2-1
\\0=(3x)^2-1^{2}$
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 find:
$(3x-1)(3x+1)=0$
Equate each factor to 0 then solve each equation to find:
$3x-1=0 \text{ or } 3x+1 = 0\\\
3x=1 \text{ or } 3x=-1
\\x=\frac{1}{3} \text{ or }x=-\frac{1}{3}$
The solution set is $\left\{-\frac{1}{3}, \frac{1}{3}\right\}$.
