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

Jul 13, 2016

$x = 3$

#### Explanation:

Squaring the bracket gives:

$y = 2 \left({x}^{2} - 6 x + 9\right)$

$y = 2 {x}^{2} - 12 x + 18$
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Using $y = a {x}^{2} + b x + c$ where

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$a = 2 \text{; "b=-12"; } c = 18$
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$x = \frac{+ 12 \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(2\right) \left(18\right)}}{2 \left(2\right)}$

$x = \frac{+ 12 \pm \sqrt{144 - 144}}{4}$

$x = + 3 \text{ }$ thus the x-axis is tangential to the vertex