Question

How do you find the roots, real and imaginary, of `y=3x^2 -11x + 4` using the quadratic formula?

Answer 1

`(11 +- sqrt73)/6`

Explanation:

`y = 3x^2 - 11x + 4 = 0`
Use the improved quadratic formula:
`D = d^2 = b^2 - 4ac = 121 - 48 = 73` --> `d = +- sqrt73`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = 11/6 +- sqrt73/6 = (11 +- sqrt73)/6`