If the area of a rectangular field is $2,\!000\,\mathrm{m^2}$ and it’s $10\,\mathrm{m}$ longer than it is wide, how long is the field? Questions like this were of interest to the ancient Babylonians. Show assumed knowledge

The area of a rectangle is width times length.
The solution of $ax^2 + bx+c=0$ is \[x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
(Section 7.5.5: Quadratic Equations).

We might decide to call the length and width $x$ and $y$ respectively ($l$ and $w$ would also be good choices for the variable names). Then the information we have is: \begin{align*} xy &= 2000\tag{$1$}\\ x &= y + 10\tag{$2$} \end{align*} We can substitute equation 2 into equation 1, obtaining: \begin{align*} (y + 10)y &= 2000\\ y^2 + 10y - 2000 &= 0\\ y &= \frac{-10\pm\sqrt{100 -4\times(-2000)}}{2}\\[5pt] &= \frac{-10\pm\sqrt{8100}}{2}\\[5pt] &= \frac{-10\pm 90}{2}\\[5pt] &= -50,40 \end{align*} A width can’t be negative, so we can discard the first solution. If $y=40$ then $x=40+10=50$ so the field is $50\,\mathrm{m}$ long.

Alternatively, you might have solved equation 2 for $y$ so we can get rid of the $y$ in the first equation and directly find $x$, the length. \begin{align*} y &= x - 10\\ x(x-10) &= 2000\\ x^2 - 10x - 2000 &= 0\\ x &= \frac{10 \pm\sqrt{100 - 4\times(-2000)}}{2}\\[5pt] &= \frac{10\pm\sqrt{8100}}{2}\\[5pt] &= \frac{10\pm 90}{2}\\[5pt] &= -40,50 \end{align*} Again we can discard the negative solution, so the length of the field is $50\,\mathrm{m}$.

It’s also possible that you solved equation 1 for $y$: \[y = \frac{2000}{x}\] and then substituted that into equation 2: \[x = \frac{2000}{x} + 10\] then multiplied both sides by $x$ to get rid of the fraction \begin{align*} x^2 &= 2000 + 10x\\ x^2 - 10x - 2000 &= 0 \end{align*} and we end up with the same quadratic equation we obtained in the second method.