How do you solve $0=x^2-3x-6$ by completing the square?

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Answer 1 · 2015-05-17T07:33:03

$0 = x^2-3x-6 = (x^2-3x+9/4)-9/4-6$

$=(x-3/2)^2-33/4$

In general:

$ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))$

Adding $33/4$ to both sides we get

$(x-3/2)^2 = 33/4$

So $x-3/2 = +-sqrt(33/4) = +-sqrt(33)/2$

Adding $3/2$ to both sides we get

$x = 3/2+-sqrt(33)/2 = (3+-sqrt(33))/2$

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