How do you solve $3x^2-4x-5=0$ by completing the square?

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Answer 1 · 2015-04-30T00:24:50

First, divide by a = 3
$x^2 - 4x/3 - 5/3 = 0$
$x^2 - 4x/3 + 4/9 - 4/9 - 5/3 = 0$

$(x - 2/3)^2 = 4/9 + 5/3 = 19/9$

a. $(x - 2/3) = (sqr19)/3$ ---> $x = 2/3 + (sqr19)/3$

b. $(x - 2/3) =(-sqr19)/3 -> x = 2/3 -(sqr19)/3$

How do you solve 3x^2-4x-5=03x^2-4x-5=0 by completing the square?