Answer:
[tex]\huge\boxed{x=\pm4.5}[/tex]
Step-by-step explanation:
The quadratic formula of
[tex]ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
We have:
[tex]y=4x^2-81\to 4x^2-81=0\\\\a=4;\ b=0;\ c=-81[/tex]
substitute
[tex]x=\dfrac{-0\pm\sqrt{0^2-4(4)(-81)}}{2(4)}=\dfrac{\pm\sqrt{1296}}{8}=\dfrac{\pm36}{8}=\pm4.5[/tex]
