## Algebra 1: Common Core (15th Edition)

$\boxed{4x +\frac{1}{3}}$
Solve the quadratic equation by using $(-b \pm \sqrt {b^2-4ac})$ $\div$ ($2a$) in $ax^2 + bx + c = 0$. $a = 6$ $b = 1$ $c = -2$ $x = $$(-1 \pm \sqrt{1 + 48}) \div 12 x =$$(-1 \pm 7)$ $\div$ $12$ The two solutions of $x$ is $\frac{1}{2}$ and $-\frac{2}{3}$. The two sides of the rectangle are $(x-\frac{1}{2})$ and $(x+\frac{2}{3})$ The perimeter of a rectangle is $2(s_1 + s_2)$ where $s_1$ is side 1 and $s_2$ is side 2. The perimeter is: $2((x-\frac{1}{2}) +(x + \frac{2}{3}))$ $2(2x + \frac{1}{6})$ $\boxed{4x +\frac{1}{3}}$