1. **State the problem:** We have an L-shaped polygon made of two rectangles with side lengths involving $x$. The area is given as 10 m², and the equation for $x$ is $$3x^2 + 2x - 1 = 0.$$ We need to find the perimeter of the shape. 2. **Recall the formula for perimeter:** The perimeter is the total distance around the shape, which is the sum of all outer side lengths. 3. **Identify the sides:** From the description: - Left vertical side: $2x + 4$ - Top horizontal side: $2$ - Bottom horizontal side: $3x + 2$ - Right vertical side: $2x$ Since the shape is L-shaped, the perimeter includes these four sides plus the missing vertical and horizontal segments that complete the shape. 4. **Find the missing sides:** The vertical segment on the right is $2x$, and the vertical segment on the left is $2x + 4$. The difference between these gives the vertical segment inside the L-shape: $$ (2x + 4) - 2x = 4 $$ Similarly, the horizontal segments are $2$ (top) and $3x + 2$ (bottom). The difference is: $$ (3x + 2) - 2 = 3x $$ 5. **Calculate the perimeter:** The perimeter is the sum of all outer sides: $$ P = (2x + 4) + 2 + (3x + 2) + 2x $$ Simplify: $$ P = 2x + 4 + 2 + 3x + 2 + 2x = (2x + 3x + 2x) + (4 + 2 + 2) = 7x + 8 $$ 6. **Solve for $x$ using the quadratic equation:** Given: $$ 3x^2 + 2x - 1 = 0 $$ Use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $a=3$, $b=2$, $c=-1$. Calculate the discriminant: $$ \Delta = 2^2 - 4 \times 3 \times (-1) = 4 + 12 = 16 $$ Calculate $x$: $$ x = \frac{-2 \pm \sqrt{16}}{2 \times 3} = \frac{-2 \pm 4}{6} $$ Two solutions: $$ x_1 = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3} $$ $$ x_2 = \frac{-2 - 4}{6} = \frac{-6}{6} = -1 $$ 7. **Choose the valid solution:** Since lengths must be positive, $x = \frac{1}{3}$ m. 8. **Calculate the perimeter:** $$ P = 7x + 8 = 7 \times \frac{1}{3} + 8 = \frac{7}{3} + 8 = \frac{7}{3} + \frac{24}{3} = \frac{31}{3} \approx 10.33 \text{ m} $$ **Final answer:** The perimeter of the shape is $$\boxed{\frac{31}{3} \text{ m} \approx 10.33 \text{ m}}.$$