How do you solve $x^2 - 6x + 4 = 0$ using the quadratic formula?

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Answer 1 · 2016-03-03T16:00:53

$x = 3 +- sqrt5$

$y = x^2 - 6x + 4 = 0$
$D = d^2 = b^2 - 4ac = 36 - 16 = 20$ --> $d = +- 2sqrt5$
$x = -b/(2a) +- d/(2a) = 6/2 +- (2sqrt5)/2 = 3 +- sqrt5$

How do you solve x^2 - 6x + 4 = 0x^2 - 6x + 4 = 0 using the quadratic formula?