# How do you find the roots, real and imaginary, of y=3x^2 - 6x + 3 using the quadratic formula?

Apr 12, 2016

Single root at $x = 1$ (multiplicity $2$)

#### Explanation:

For a quadratic in the general form: $y = a {x}^{2} + b x + c$
the roots are given by the quadratic formula as
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Given: $y = 3 {x}^{2} - 6 x + 3$
$\textcolor{w h i t e}{\text{XXX}} a = 3$
$\textcolor{w h i t e}{\text{XXX}} b = - 6$
$\textcolor{w h i t e}{\text{XXX}} c = 3$

So the roots are
color(white)("XXX")x=(6+-sqrt((-6)^2-4(3)(3)))/(2(3)

$\textcolor{w h i t e}{\text{XXXX}} = \frac{6 \pm \sqrt{36 - 36}}{6}$

$\textcolor{w h i t e}{\text{XXXX}} = 1$