How do you find the zeros, real and imaginary, of #y=5x^2-x+12# using the quadratic formula?

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Answer 1 · 2017-01-09T18:29:33

The equation has #2# complex zeros. See explanation.

First we have to calculate the discriminant:

#Delta=b^2-4ac#

#Delta=(-1)^2-4xx5xx12=1-240=-239#

The discriominant is negative, so the equation has 2 complex zeros:

#x_1=(-b-sqrt(Delta))/(2a)=(1-sqrt(-239))/10=0.1-0.1*sqrt(239)i#

#x_2=(-b+sqrt(Delta))/(2a)=(1+sqrt(-239))/10=0.1+0.1*sqrt(239)i#