## Algebra 2 Common Core

The solutions are $-\frac{5}{2}$ and $\frac{5}{2}$.
Write each term of the binomial in square form to get $(2x)^2-5^2=0$. Factor the difference of two squares to get $(2x-5)(2x+5)=0.$ Equate each factor to 0, then solve each equation for x: $2x-5=0 \text{ or } 2x+5=0 \\x=\frac{5}{2} \text{ or } x=-\frac{5}{2}.$ Thus, the solutions are $-\frac{5}{2}$ and $\frac{5}{2}$.