#### Answer

$\boxed{4x +\frac{1}{3}}$

#### Work Step by Step

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})) $
$2(2x + \frac{1}{6})$
$\boxed{4x +\frac{1}{3}}$