Question

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

Answer 1

The zeros are `x=-0.5+i5.895` and `x=-0.5-i5.895`.

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
`ax^2+bX+c=0` has solutions `x=(-b\pmsqrt(b^2-4ac))/(2a)`.

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

`x=(-1\pmsqrt(1^2-4*1*35))/(2)`
`=(-1\pmsqrt(-139))/2`
We notice that the root is negative and then imaginary. So I rewrite it using the fact that `i=sqrt(-1)`

`x=(-1\pmisqrt(139))/2\approx-1/2\pmi11.79/2=-0.5\pmi5.895`

The two solutions are then `x=-0.5+i5.895` and `x=-0.5-i5.895`.