# How do you find the zeros, real and imaginary, of y=- x^2-22x+6  using the quadratic formula?

Apr 10, 2016

Zeros are $- 11 + \sqrt{127}$ and $- 11 - \sqrt{127}$

#### Explanation:

The roots of general form of equation $a {x}^{2} + b x + c = 0$ are zeros of the general form of equation $y = a {x}^{2} + b x + c$.

In the given equation $y = - {x}^{2} - 22 x + 6$, we see that $a = - 1$, $b = - 22$ and $c = 6$.

As discriminant ${b}^{2} - 4 a c = {\left(- 22\right)}^{2} - 4 \left(- 1\right) \left(6\right) = 484 + 24 = 508$

As discriminant is positive but not a complete square, we have real but irrational roots.

Hence, using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$,

the zeros are the given equation are $x = \frac{- \left(- 22\right) \pm \sqrt{508}}{2 \cdot \left(- 1\right)}$

or $x = - \frac{22 \pm \sqrt{508}}{2} = - \frac{22 \pm 2 \sqrt{127}}{2}$ i.e.

zeros are $- 11 + \sqrt{127}$ and $- 11 - \sqrt{127}$