Question

How do you find the zeros of `y = 2x^2 + 4x -1` using the quadratic formula?

Answer 1

`-1+-sqrt(6)/2`

Explanation:

This quadratic function is written in "standard form", or `ax^2+bx+c`.
The quadratic formula uses the numbers a, b, and c like this:
`(-b+-sqrt(b^2-4ac))/(2a)`
comparing the original function to the general, standard form function above, we see that `a=2, b=4, &c=-1`.

Plug these in the quadratic formula:
`(-4+-sqrt(4^2-4(2)(-1)))/(2*2)=(-4+-sqrt(16-(-8)))/4`
`(-4+-sqrt(24))/4`
To simplify that `sqrt(24)`, pull out the largest perfect square that goes in it. Since 4 divides evenly into 24, and neither 9 nor 16 do, break it down in terms of 4: `sqrt(24)=sqrt(4*6)`
You can take out `sqrt(4)=2`:

`sqrt(4*6)=2sqrt(6)`
From here, divide through by 4 to get `-1+-sqrt(6)/2`

...and remember, quadratic functions always have two solutions!