Question

How do you find the zeros, if any, of `y= 32x^2 -32x+ 2`using the quadratic formula?

Answer 1

`x=1/2+-sqrt(3)/4`

Explanation:

The quadratic formula for the zeros of the general equation
`color(white)("XXX")y=ax^2+bx+c`
is given by
`color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)`

For the given equation
`color(white)("XXX")y=3x^2-32x+2`
this becomes
`color(white)("XXX")(32+-sqrt(32^2-4(32)(2)))/(2(32)`

and from there on, it's just basic arithmetic to get
`color(white)("XXX")x=1/2+-sqrt(3)/4`