How do you solve $6x^2-5x+3=0$ ?

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Answer 1 · 2015-05-23T12:35:42

y = 6x^2 - 5x + 3 = 0

D = d^2 = b^2 - 4ac = 25 - 72 = -47 < 0. There are no real roots, meaning the parabola graph doesn't intersect the x-axis.
There are 2 complex roots:

$d = +- sqrt47i$

$x = 5/12 +- (sqrt47i)/12$

How do you solve 6x^2-5x+3=06x^2-5x+3=0?