## Elementary Algebra

The solution set is $\left\{-\frac{3}{2}, \frac{3}{2}\right\}$.
Add $-9$ to both sides to obtain: $4x^2-9=0 \\(2x)^2-3^2=0$ 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: $(2x-3)(2x+3)=0$ Equate each factor to 0, and then solve each equation to obtain: $2x-3=0 \text{ or } 2x+3 = 0 \\2x=3 \text{ or } 2x=-3 \\x=\frac{3}{2} \text{ or } x=-\frac{3}{2}$ The solution set is $\left\{-\frac{3}{2}, \frac{3}{2}\right\}$.