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

1 Answer
Nghi N.
Apr 26, 2016

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

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