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

Aug 17, 2017

You gotta re-write it in the proper form...

#### Explanation:

Remember that your quadratic equation is meant to find x for equations of form

$a {x}^{2} + b x + c = 0$

So you start by:

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

$= - 2 \left({x}^{2} - 4 x + 4\right) - {x}^{2} + x + 12$

$= - 2 {x}^{2} + 8 x - 8 - {x}^{2} + x + 12$

$= - 3 {x}^{2} + 9 x + 4$

...so now you have your coefficients a, b, and c for the quadratic equation.

$a = - 3 , b = 9 , c = 4$

Plug these in to the quadratic formula:

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

I get roughly $x = 3.393 \mathmr{and} - 0.393$

It's always good to check your answers by plugging these values for x into your original equation. You should get zero. (which you don't, with these values, exactly, because I've rounded off. But it's very close.)

GOOD LUCK