# How do you find the roots, real and imaginary, of y=-(x -2)^2-4x+5 using the quadratic formula?

Jul 13, 2017

$\left\{- 1 , 1\right\}$

#### Explanation:

We have roots of the equation and zeros of the function.

Here $y = - {\left(x - 2\right)}^{2} - 4 x + 5$

= $- {x}^{2} + 4 x - 4 - 4 x + 5$

= $- {x}^{2} + 1$

According ro quadratic formula, zeros of function $f \left(x\right) = a {x}^{2} + b x + c$ are $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Here $a = - 1$, $b = 0$ and $c = 1$

hence zeros are $\frac{\pm \sqrt{- 4 \times \left(- 1\right) \times 1}}{- 2} = \pm \frac{2}{- 2}$ or $1 , - 1$