How do you solve using the completing the square method $2 x^{2} + 3 x - 5 = 0$ $2x^2+3x-5=0$ ?

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Answer 1 · 2018-03-03T07:16:41

$x = \pm \frac{7}{4} - \frac{3}{4} = 1, - \frac{5}{2}$ $color(brown)(x = +-7/4 - 3/4 = 1, -5/2$

$2 x^{2} + 3 x - 5 = 0$ $2x^2 + 3x - 5 = 0$

$cancel2(x^2 + (3/2)x - 5/2 )= 0$

$(x^2 + (2 * (3/4) * x ) = 5/2$

Add $(3/4)^2$ to both sides.

$x^2 + (2 * (3/4) * x) + (3/4)^2 = 5/2 + (3/4)^2 = 49/16$

$(x + 3/4)^2 = (sqrt(49/16))^2 = (7/4)^2$

$x + 3/4 = +- 7/4$

$color(brown)(x = +-7/4 - 3/4 = 1, -5/2$

How do you solve using the completing the square method 2x^2+3x-5=02x2+3x−5=0 2x^2+3x-5=0?