# How do you find the zeros, real and imaginary, of y=-x^2+7x-11 using the quadratic formula?

Mar 8, 2016

Zeros are $\frac{7}{2} - \frac{\sqrt{5}}{2}$ and $\frac{7}{2} + \frac{\sqrt{5}}{2}$

#### Explanation:

Zeros, real and imaginary, of $y = a {x}^{2} + b x + c$ are given by the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In y=−x^2+7x−11, $a = - 1$, $b = 7$ and $c = - 11$

Hence zeros are given by

$\frac{- 7 \pm \sqrt{{7}^{2} - 4 \times \left(- 1\right) \times \left(- 11\right)}}{2 \times \left(- 1\right)}$ or

$\frac{- 7 \pm \sqrt{49 - 44}}{- 2}$

$\frac{- 7 \pm \sqrt{5}}{- 2}$ or

$\frac{7}{2} \pm \frac{\sqrt{5}}{2}$

Hence zeros are $\frac{7}{2} - \frac{\sqrt{5}}{2}$ and $\frac{7}{2} + \frac{\sqrt{5}}{2}$