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

1 Answer
Jan 1, 2016

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.

Now use the quadratic formula:

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.