How do you solve using the quadratic formula $3x(x - 2) = 1$ ?

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Answer 1 · 2015-05-12T03:04:36

f(x) = 3x^2 - 6x - 1 = 0

$D = d^2 = b^2 - 4ac = 36 + 12 = 48 = 16.3 -> d = +-4.sqr3$

$x = 6/6 + (4.sqr3)/6$ = $1 + (sqr3)/3$

$x = 6/6 - (sqr3)/3 = 1 - (sqr3)/3$

How do you solve using the quadratic formula 3x(x - 2) = 13x(x - 2) = 1?