How do you solve by completing the square for $3 x^{2} = 6 x + 4$ $3x^2 = 6x + 4$ ?

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Answer 1 · 2015-05-23T00:41:21

$f (x) = 3 x^{2} - 6 x - 4$ $f(x) = 3x^2 - 6x - 4$ = 0

f(x) = 3(x^2 - 2x - 4/3) = 3(x^2 - 2x + 1) - 3 - 12/3

(x - 1)^2 = 21/9 =

$(x - 1) = \pm \frac{\sqrt{21}}{3}$ $(x - 1) = +- (sqrt21)/3$ ->

$(x - 1) = \frac{\sqrt{21}}{3} \to x = 1 + \frac{\sqrt{21}}{3}$ $(x - 1) = (sqrt21)/3 -> x = 1 + (sqrt21)/3$

$(x - 1) = - \frac{\sqrt{21}}{3} \to x = 1 - \frac{\sqrt{21}}{3}$ $(x - 1) = -(sqrt21)/3 -> x = 1 - (sqrt21)/3$

How do you solve by completing the square for 3x2=6x+43x^2 = 6x + 4?