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

May 12, 2015

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

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

$x = \frac{6}{6} + \frac{4. s q r 3}{6}$= $1 + \frac{s q r 3}{3}$

$x = \frac{6}{6} - \frac{s q r 3}{3} = 1 - \frac{s q r 3}{3}$