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

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Answer 1 · 2016-07-02T20:14:48

The polynomial is of order 2 so has two roots. These are real, distinct and given by:

#x = -1/6 - (sqrt(13))/6# or #x = -1/6 + (sqrt(13))/6#

The quadratic formula, for an equation of the form #ax^2 + bx + c = 0# is given by:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

For this question a = -6, b = -2, c = 2.

#x = (2 +- sqrt(4 - 4(-6)(2)))/(2(-6))#

#x = (2 +- sqrt(52))/(-12) = (2 +- 2sqrt(13))/(-12)#

#x = -1/6 - (sqrt(13))/6# or #x = -1/6 + (sqrt(13))/6#