Answer:
[tex]A=(2x^{2}+9x-18)\ m^{2}[/tex]
Step-by-step explanation:
Let
L-----> the length of the concert hall
W----> the width of the concert hall
we know that
The area of a rectangle ( concert hall) is equal to
[tex]A=LW[/tex]
we have
[tex]L=(x+6)\ m[/tex]
[tex]W=(2x-3)\ m[/tex]
substitute the values
[tex]A=(x+6)(2x-3)\\ \\A=2x^{2} -3x+12x-18\\ \\A=(2x^{2}+9x-18)\ m^{2}[/tex]