# How do you find the zeros, real and imaginary, of y=- x^2+12x+6  using the quadratic formula?

Jan 13, 2018

Solution ; Zeros are two real number $x \approx 12.6 , x \approx - 0.48$

#### Explanation:

$y = - {x}^{2} + 12 x + 6$ Comparing with standard quadratic

equation $a {x}^{2} + b x + c = 0$ and $a$ is not zero. Here

a=-1 . b= 12 ,c = 6 ; D is discriminant and $D = {b}^{2} - 4 a c$

$\therefore D = {b}^{2} - 4 a c = {12}^{2} - 4 \cdot \left(- 1\right) \cdot 6 = 168$ . If $D$ is positive,

we get two real solutions, if it is zero just one solution, and

if it is negative we get two complex (imaginary) solutions.

Quadratic Formula: $x = \frac{- b \pm \sqrt{\left({b}^{2} - 4 a c\right)}}{2 a}$

$\therefore x = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \cdot \left(- 1\right) \cdot 6}}{2 \cdot \left(- 1\right)}$ or

$x = \frac{- 12 \pm \sqrt{168}}{- 2} \approx 6 \pm 6.48$ or

$x \approx 12.48 \left(2 \mathrm{dp}\right) \mathmr{and} x \approx - 0.48 \left(2 \mathrm{dp}\right)$

Solution ; Zeros are two real number $x \approx 12.6 , x \approx - 0.48$

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