Question

How do you find the zeros, real and imaginary, of `y=-4x^2-2x+3`using the quadratic formula?

Answer 1

`(-1+--1sqrt(13))/4`

Explanation:

The quadratic formula is `-b/(2a)+-sqrt(b^2-4*a*c)/(2a)`

Let's look at our equation, `-4x^2-2x+3`, in a more general form: `ax^2-bx+c`.

So `a` is `-4`
`b` is `-2`
`c` is `3`.

Now let's plug our variables into the quadratic formula. I like to put parentheses around the variables I'm replacing, just to be extra careful about silly little sign errors.

`(-(-2))/(2(-4))+-sqrt((-2)^2-4*(-4)*(3))/(2(-4))`.

This becomes `2/-8+-sqrt(4-(-48))/-8`, which we can simplify to `-1/4+-(2sqrt(13))/-8`, or `-1/4+-(-1sqrt(13))/4`.

The final answer is `(-1+--1sqrt(13))/4`