Question

How do you find the zeros, real and imaginary, of `y= -7x^2-3x+2` using the quadratic formula?

Answer 1

About `-0.79` and `0.36`.

Explanation:

This quadratic is in standard form: `ax^2+bx+c`

So for `y=-7x^2-3x+2`,
`a=-7`,
`b=-3`, and
`c=2`.

Now we use the quadratic formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

and plug in our `a`, `b`, and `c` values and simplify:

`x=(3+-sqrt((-3)^2-4(-7)(2)))/(2(-7))`

`=(3+-sqrt(9+56))/(-14)`

`=(3+-sqrt(65))/(-14)`

which splits into the two solutions

`x=(3+sqrt(65))/(-14) ~~ -0.79`
and
`x=(3-sqrt(65))/(-14) ~~ 0.36`