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

Jan 1, 2016

First, equate to $0$

ie. $3 {x}^{2} - 2 x + 12 = 0$

This is of the form $a {x}^{2} + b x + c$, where $a = 3$ , $b = - 2$ and $c = 12$.

x=(-b±sqrt(b^2-4ac))/(2a)
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 $\frac{1}{3} + \frac{5}{3} i$ and $\frac{1}{3} - \frac{5}{3} i$.