Question

How do you find the zeros, real and imaginary, of `y= 7x^2-6x+14` using the quadratic formula?

Answer 1

By plugging in the coefficients and constant as a, b, and c into the quadratic formula.

Explanation:

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

The a, b, and c refers to a, b, and c in standard form which is `ax^2+bx+c`. So given the equation `y=7x^2-6x+14, a=7, b=-6, c=14`

Now we begin the process of plugging into the formula.

`(-(-6)+-sqrt((-6)^2-4(7)(14)))/(2*(7))`
`(6+-sqrt(36-392))/14`

Now at this point, I can stop because I know all the answers are imaginary, we know this because under the square root, there will be a negative number when we take 392 from 36. Because of this, the answers are imaginary.