## Elementary Algebra

$\left\{-\frac{5}{4}\right\}$
The equation can be written as: $(4y)^2+2(4y)(5) + 5^2=0$ The left side of the equation is a perfect square trinomial of the form $a^2+2ab+b^2$ where $a=4y$ and $b=5$. RECALL: $a^2+2ab+b^2=(a+b)^2$ Factor the trinomial using the formula above with $a=4y$ and $b=5$ to obtain: $(4y+5)^2=0$ Equate the factor to 0 to obtain: $4y+5=0$ Solve the equation to obtain: $4y=-5 \\y=\frac{-5}{4}$ Therefore the solution set is $\left\{-\frac{5}{4}\right\}$.