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:

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

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