# How do you find the zeros of  y = -x^2 - 9x + 3  using the quadratic formula?

Jul 21, 2016

$- \frac{9}{2} - \frac{\sqrt{93}}{2}$ and $- \frac{9}{2} + \frac{\sqrt{93}}{2}$,

are real zeros of $y = - {x}^{2} - 9 x + 3$.

#### Explanation:

Zeros of $y = - {x}^{2} - 9 x + 3$ are those values of $x$ for which $y = 0$. Hence, zeros are given by the solution of quadratic equation $- {x}^{2} - 9 x + 3 = 0$ or ${x}^{2} + 9 x - 3 = 0$.

Quadratic formula gives solution of equation $a {x}^{2} + b x + c = 0$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. Hence for ${x}^{2} + 9 x - 3 = 0$, we have

$x = \frac{- 9 \pm \sqrt{{9}^{2} - 4 \cdot 1 \cdot \left(- 3\right)}}{2 \cdot 1}$

= $\frac{- 9 \pm \sqrt{81 + 12}}{2}$

= $\frac{- 9 \pm \sqrt{93}}{2}$

= $- \frac{9}{2} - \frac{\sqrt{93}}{2}$ and $- \frac{9}{2} + \frac{\sqrt{93}}{2}$,

are real zeros of $y = - {x}^{2} - 9 x + 3$.