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

Question

1414 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-19T17:25:44

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

$[x^2 + 2x - 4/3 = x^2 + 2x + 1 - 1 - 4/3 = 0$

$y = (x + 1)^2 - 7/3 = 0$

$(x + 1)^2 = 7/3$ -> $(x + 1 )= +-(sqrt7)/(sqrt3)$

$x = -1 +- (sqrt21/3)$

How do you solve 3x^2 + 5x = -x + 43x^2 + 5x = -x + 4 by completing the square?