# How do you find the zeros, real and imaginary, of y=-x^2- x -35 using the quadratic formula?

May 30, 2016

The zeros are $x = - 0.5 + i 5.895$ and $x = - 0.5 - i 5.895$.

#### Explanation:

The zeros are the solution of this equations

$- {x}^{2} - x - 35 = 0$

first of all we can remove all the minus multiplying left and right for $- 1$, but on the right there is zero that does not change sign.

${x}^{2} + x + 35 = 0$

The quadratic formula tells us that an equation in the form
$a {x}^{2} + b X + c = 0$ has solutions $x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

For us $a = 1$, $b = 1$ and $c = 35$, I substitute

$x = \frac{- 1 \setminus \pm \sqrt{{1}^{2} - 4 \cdot 1 \cdot 35}}{2}$
$= \frac{- 1 \setminus \pm \sqrt{- 139}}{2}$
We notice that the root is negative and then imaginary. So I rewrite it using the fact that $i = \sqrt{- 1}$

$x = \frac{- 1 \setminus \pm i \sqrt{139}}{2} \setminus \approx - \frac{1}{2} \setminus \pm i \frac{11.79}{2} = - 0.5 \setminus \pm i 5.895$

The two solutions are then $x = - 0.5 + i 5.895$ and $x = - 0.5 - i 5.895$.