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

1 Answer
Mar 7, 2016

Answer:

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.