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

Dec 13, 2017

$x = \frac{7 + \sqrt{93}}{2} \mathmr{and} \frac{7 - \sqrt{93}}{2}$

Explanation:

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

Putting our values in gives us:
$x = \frac{- 7 \pm \sqrt{{7}^{2} - 4 \left(- 1 \cdot 11\right)}}{- 2}$

$= \frac{- 7 \pm \sqrt{49 - 4 \left(- 11\right)}}{- 2}$

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

$= \frac{- 7 \pm \sqrt{93}}{-} 2$

$= \frac{- 7 - \sqrt{93}}{-} 2 \mathmr{and} \frac{- 7 + \sqrt{93}}{-} 2$

$= \frac{7 + \sqrt{93}}{2} \mathmr{and} \frac{7 - \sqrt{93}}{2}$