Hello, archistrategos214!
Did you make a sketch?
A tunnel has a height of 20m and a width of 36m at the base.
If the vertex is at the top of the arch, at which point above the base is it 18m wide? Code:
20
* * *
*  *
*  *
*  *

 *      +      * 
18  18
The vertex of the parabola is on the yaxis.
. . The equation has the form: .$\displaystyle y \:=\:ax^2 + c$
Since $\displaystyle (0,20)$ is on the parabola: .$\displaystyle 20 \:=\:a\!\cdot\!0^2 + c\quad\Rightarrow\quad c = 20$
. . The equation (so far) is: .$\displaystyle y \:=\:ax^2 + 20$
Since $\displaystyle (18,0)$ is on the parabola: .$\displaystyle 0 \:=\:a\!\cdot\!18^2 + 20\quad\Rightarrow\quad a = \frac{5}{81}$
. . The parabola is: .$\displaystyle y \:=\:\frac{5}{81}x^2 + 20$
"Where is the parabola 18m wide?" $\displaystyle \rightarrow$ "Where is $\displaystyle x = 9$ ?"
. . When $\displaystyle x = 9\!:\;\;y \;=\;\frac{5}{81}(9^2) + 20 \;=\;15$
The arch is 18m wide at a point 15m above the base.