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

Feb 3, 2016

$x = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$

#### Explanation:

The quadratic formula for the zeros of the general equation
$\textcolor{w h i t e}{\text{XXX}} y = a {x}^{2} + b x + c$
is given by
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

For the given equation
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} - 32 x + 2$
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
$\textcolor{w h i t e}{\text{XXX}} x = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$