# How do you find the zeros, real and imaginary, of y=3x^2-6x+9 using the quadratic formula?

the zeros are
${x}_{1} = 1 + \sqrt{2} i$
${x}_{2} = 1 - \sqrt{2} i$

#### Explanation:

In order to solve for the zeros, we equate the given equation to zero.

Let y=0
$3 {x}^{2} - 6 x + 9 = 0$
divide both sides of the equation by 3
${x}^{2} - 2 x + 3 = 0$
We now take note of the numerical coefficients of each term.

$1 \cdot {x}^{2} + \left(- 2\right) \cdot x + 3 = 0$

The quadratic equation is given by
$a {x}^{2} + b x + c = 0$
We can see clearly that $a = 1$ and $b = - 2$ and $c = 3$

The quadratic formula for solving the unknown x is given by

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

direct substitution

$x = \frac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$x = \frac{2 \pm \sqrt{4 - 12}}{2}$

$x = \frac{2 \pm \sqrt{- 8}}{2}$

$x = \frac{2 \pm 2 \sqrt{2} i}{2}$

$x = 1 \pm \sqrt{2} i$

The zeros are
${x}_{1} = 1 + \sqrt{2} i$
${x}_{2} = 1 - \sqrt{2} i$

Check at ${x}_{1} = 1 + \sqrt{2} i$ using the original equation

$3 {x}^{2} - 6 x + 9 = 0$
$3 {\left(1 + \sqrt{2} i\right)}^{2} - 6 \left(1 + \sqrt{2} i\right) + 9 = 0$
$3 \left(1 + 2 \sqrt{2} i + {\left(\sqrt{2}\right)}^{2} \cdot {i}^{2}\right) - 6 - 6 \sqrt{2} i + 9 = 0$
$3 \left(1 + 2 \sqrt{2} i + \left(2\right) \cdot \left(- 1\right)\right) - 6 - 6 \sqrt{2} i + 9 = 0$

$3 \left(1 + 2 \sqrt{2} i - 2\right) - 6 - 6 \sqrt{2} i + 9 = 0$

$3 \left(2 \sqrt{2} i - 1\right) - 6 - 6 \sqrt{2} i + 9 = 0$

$6 \sqrt{2} i - 3 - 6 - 6 \sqrt{2} i + 9 = 0$

$0 = 0$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check at ${x}_{1} = 1 - \sqrt{2} i$ using the original equation

$3 {x}^{2} - 6 x + 9 = 0$
$3 {\left(1 - \sqrt{2} i\right)}^{2} - 6 \left(1 - \sqrt{2} i\right) + 9 = 0$
$3 \left(1 - 2 \sqrt{2} i + {\left(- \sqrt{2}\right)}^{2} \cdot {i}^{2}\right) - 6 + 6 \sqrt{2} i + 9 = 0$
$3 \left(1 - 2 \sqrt{2} i + \left(2\right) \cdot \left(- 1\right)\right) - 6 + 6 \sqrt{2} i + 9 = 0$

$3 \left(1 - 2 \sqrt{2} i - 2\right) - 6 + 6 \sqrt{2} i + 9 = 0$

$3 \left(- 2 \sqrt{2} i - 1\right) - 6 + 6 \sqrt{2} i + 9 = 0$

$- 6 \sqrt{2} i - 3 - 6 + 6 \sqrt{2} i + 9 = 0$

$0 = 0$

God bless...I hope the explanation is useful.