Question

Using the quadratic formula y=4x ²-81

Answer 1

`\huge\boxed{x=\pm4.5}`

Explanation:

The quadratic formula of

`ax^2+bx+c=0\\\\x=(-b\pm√(b^2-4ac))/(2a)`

We have:

`y=4x^2-81\to 4x^2-81=0\\\\a=4;\ b=0;\ c=-81`

substitute

`x=(-0\pm√(0^2-4(4)(-81)))/(2(4))=(\pm√(1296))/(8)=(\pm36)/(8)=\pm4.5`