Answer:
[tex]x=\pm\frac{9}{2}[/tex]
Step-by-step explanation:
I assume you mean [tex]4x^2-81=0[/tex]:
[tex]4x^2-81=0\\(2x-9)(2x+9)=0\\x=\frac{9}{2},\:x=-\frac{9}{2}\\x=\pm\frac{9}{2}[/tex]
We were able to work quickly here since the original left-side expression was a difference of squares.
