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

Nov 29, 2015

Expand the expression on the right side into standard form, then apply the quadratic formula.
In this case there will be two imaginary roots $2 \pm i \sqrt{5}$

#### Explanation:

$y = {\left(x - 3\right)}^{2} + 2$

$\Rightarrow y = 1 {x}^{2} + \left(- 4\right) x + 9$

which is an example of quadratic standard form: $y = a {x}^{2} + b x + c$
for which the roots are given by the quadratic formula
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- b + 1 \sqrt{{b}^{2} - 4 a c}}{- 2 a}$

Which for our example becomes:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(9\right)}}{2 \left(1\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{4 \pm \sqrt{16 - 36}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{4 \pm 2 \sqrt{- 5}}{2}$

$\textcolor{w h i t e}{\text{XXXX}} = 2 \pm i \sqrt{5}$