Question

How do you find the roots, real and imaginary, of `y= 12x^2+3x-6` using the quadratic formula?

Answer 1

`x = -1/8+- sqrt33/8`

Explanation:

`y = 12x^2 + 3x - 6 = 4x^2 + x - 2 = 0`
`D = d^2 = b^2 - 4ac = 1 + 32 = 33` --> `d = +-sqrt33`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = -1/8 +- sqrt33/8`