1. **State the problem:** We are given a trapezium with dimensions based on $x$: the top side is $x - 4$, the bottom side is $x + 5$, and the height (left vertical side) is $2x$. 2. **Given equation:** We also have a quadratic equation related to the trapezium dimensions: $$2x^2 + x - 351 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a = 2$, $b = 1$, and $c = -351$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 1^2 - 4(2)(-351) = 1 + 2808 = 2809$$ Find the square root: $$\sqrt{2809} = 53$$ Compute the roots: $$x = \frac{-1 \pm 53}{2 \times 2} = \frac{-1 \pm 53}{4}$$ Two possible values: - $$x = \frac{-1 + 53}{4} = \frac{52}{4} = 13$$ - $$x = \frac{-1 - 53}{4} = \frac{-54}{4} = -13.5$$ 4. **Interpretation:** Since $x$ is a length, it cannot be negative. So, $x = 13$. 5. **Calculate the dimensions:** - Top side: $$13 - 4 = 9$$ - Bottom side: $$13 + 5 = 18$$ - Height: $$2 \times 13 = 26$$ 6. **Summary:** The trapezium has a height of 26 units, with top base 9 units and bottom base 18 units. --- Final answer: $$x = 13$$