# How do you find the zeros, real and imaginary, of y=- x^2-22x+67  using the quadratic formula?

Apr 28, 2016

Zeros of $y = - {x}^{2} - 22 x + 47$ are $- 11 + 2 \sqrt{67}$ and $- 11 - 2 \sqrt{47}$

#### Explanation:

To find zeros of $y = - {x}^{2} - 22 x + 67$, one needs to find values of $x$ for which $y = f \left(x\right) = 0$.

For $a {x}^{2} + b x + c = 0$, solution is given by quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. Hence for $- {x}^{2} - 22 x + 67 = 0$ or ${x}^{2} + 22 x - 67 = 0$,

$x = \frac{- 22 \pm \sqrt{{22}^{2} - 4 \times 1 \times \left(- 67\right)}}{2}$

= $\frac{- 22 \pm \sqrt{484 + 268}}{2}$

= $\frac{- 22 \pm \sqrt{752}}{2}$

= $\frac{- 22 \pm 4 \sqrt{47}}{2}$

= $- 11 \pm 2 \sqrt{47}$