We have the following:
[tex]\begin{gathered} A=w\cdot l \\ A=6x^{2}+x-1 \\ w=2 \end{gathered}[/tex]to calculate the length:
[tex]\begin{gathered} l=\frac{A}{w} \\ l=\frac{6x^{2}+x-1}{2} \\ l=\frac{6x^2}{2}+\frac{x}{2}-\frac{1}{2} \\ l=3x^2+\frac{x}{2}-\frac{1}{2} \end{gathered}[/tex]the perimeter is:
[tex]p=2\cdot w+2\cdot l[/tex]replacing:
expanded expression:
[tex]\begin{gathered} p=2\cdot2+2\cdot(3x^2+\frac{x}{2}-\frac{1}{2}) \\ p=4+2\cdot(3x^2+\frac{x}{2}-\frac{1}{2}) \end{gathered}[/tex]simplifed expression:
[tex]\begin{gathered} p=2\cdot2+2\cdot(3x^2+\frac{x}{2}-\frac{1}{2}) \\ p=4+2\cdot3\cdot x^2+2\cdot\frac{x}{2}-2\cdot\frac{1}{2} \\ p=4+6x^2+x-1 \\ p=6x^2+x+3 \end{gathered}[/tex]