Given
The base and height of a parallelogram is (x-7)meters and (x+9)meters respectively.

Formula


We now substitute into the formula

It is now Quadratic Equation
![\begin{gathered} a=1,\text{ b=2 and c=-255} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ We\text{ can now substitute into the quadratic formula} \\ x=\frac{-2\pm\sqrt[]{2^2^{}-4*1*-255}}{2*1} \\ \\ x=\frac{-2\pm\sqrt[]{4^{}--1020}}{2} \\ \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/e1juu2es6ngrjuqywkhwh8tvisyts34h2u.png)
![\begin{gathered} x=\frac{-2\pm\sqrt[]{1024}}{2} \\ \\ x=(-2\pm32)/(2) \\ \\ x=(-2+32)/(2)=(30)/(2)=15 \\ \\ or \\ x=(-2-32)/(2)=-(34)/(2)=-17 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wforcgtwdrluv0vnlhfb12iqwo2nxhdmjj.png)
For distance or dimension it can't be negative, so we choose the positive
x=15
Recall from the question
Base=(x-7)meters
Height= (x+9)meters
We can now replace x with 15

The final answer
Base is 8m
Height is 24m