How do you solve $x^2 + 3x - (7/4) = 0$ by completing the square?

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Answer 1 · 2018-04-12T15:17:12

$x = -1.5+-2$

$x^2+3x−7/4=0$

$(x+3/2)^2 - (3/2)^2 - 7/4 = 0$

$(x+3/2)^2 - 9/4 - 7/4 = 0$

$(x+3/2)^2 - 16/4 = 0$

$(x+3/2)^2 = 4$

$x+3/2 = sqrt4$

$color(red)(x= -3/2+-2)$

$color(red)(x=0.5 or x = -3.5)$

How do you solve x^2 + 3x - (7/4) = 0x^2 + 3x - (7/4) = 0 by completing the square?