# How do you find the roots, real and imaginary, of y= x^2 - 12x -9 -2(x- 3 )^2  using the quadratic formula?

Aug 26, 2017

$x = 2 i \sqrt{3} , - 2 i \sqrt{3}$

#### Explanation:

First, let's simplify the equation.

$y = {x}^{2} - 12 x - 9 - 2 {\left(x - 3\right)}^{2} \to$

$y = {x}^{2} - 12 x - 9 - 2 \left({x}^{2} - 6 x + 9\right) \to$

$y = {x}^{2} - 12 x - 9 - 2 {x}^{2} + 12 x - 18 \to$

$y = - {x}^{2} - 27 = - 1 {x}^{2} + 0 x - 27$

I put the -1 and 0 to show what the a and b terms will be. Using the simplified version of the equation, we can find the roots using the quadratic equation. In the quadratic equation, $a = - 1 , b = 0 , \mathmr{and} c = - 27$.

x=(-b±sqrt(b^2-4ac))/(2a) ->

x=(-(0)±sqrt((0)^2-4(-1)(-27)))/(2(-1)) ->

x=(±sqrt(-(1)(4)(27)))/-2 ->

x=(±sqrt(-1)sqrt(4)sqrt(27))/-2 ->

x=(±i(2)(3sqrt(3)))/-2 ->

x=±2isqrt(3)
OR
$x = 2 i \sqrt{3} , - 2 i \sqrt{3}$