How do you find the zeros, real and imaginary, of y= 3x^2-6x+14 using the quadratic formula?

1 Answer
Nghi N.
Jul 12, 2016

1 +- (isqrt33)/3

Explanation:

y = 3x^2 - 6x + 14 = 0
D = d^2 = 36 - 168 = - 132 < 0
Since D < 0, there are no real roots. There are 2 imaginary roots.
d = +- 2isqrt33
Two imaginary roots:
x = -b/(2a) +- d/(2a) = 6/6 +- isqrt33/3 = 1 +- isqrt33/3

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