How do you find the roots, real and imaginary, of #y=3x^2-2x+12# using the quadratic formula?

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Answer 1 · 2016-01-01T15:00:56

First, equate to #0#

ie. #3x^2 - 2x +12 = 0#

This is of the form #ax^2 + bx + c#, where #a=3# , #b=-2# and #c=12#.

#x=(-b±sqrt(b^2-4ac))/(2a)#

Substitute values into formula:

#x=(-(-2)±sqrt((-2)^2-4(3)(12)))/(2(3))=(2±sqrt(4-144))/6=(2±sqrt(-100))/6=(2±10i)/6=(1pm5i)/3=1/3pm5/3i#

Roots are #1/3+5/3i# and #1/3-5/3i#.