How do you solve the equation $3x^2+6x=2x^2+3x-4$ by completing the square?

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Answer 1 · 2017-07-10T00:33:45

$x = -3/2 -sqrt(7)/2 \ i, -3/2 +sqrt(7)/2 \ i$

We have:

$3x^2+6x = 2x^2+3x-4$

First let us collect terms on the LHS

$3x^2 - 2x^2 + 6x - 3x -4 = 0$
$:. x^2 + 3x +4 = 0$

Now we complete the square, by forming a term with half the " $b$ " value; that is

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

Which we can now re-arrange and solve;

$:. (x + 3/2)^2 = -7/4$

$:. x + 3/2 = +-sqrt(7)/2 \ i$

$:. x = -3/2 +-sqrt(7)/2 \ i$

Leading to the two complex conjugate solutions:

$x = -3/2 -sqrt(7)/2 \ i, -3/2 +sqrt(7)/2 \ i$

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