Question

How do you find the roots, real and imaginary, of `y= 4x^2-3x-6` using the quadratic formula?

Answer 1

`x = (3 +- sqrt105)/8`

Explanation:

`y = 4x^2 - 3x - 6 = 0`.
Use the improved quadratic formula (Socratic Search) -->
`D = d^2 = b^2 - 4ac = 9 + 96 = 105` --> `d = +-sqrt105`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = 3/8 +- sqrt105/8 = (3 + sqrt105)/8`