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

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Answer 1 · 2015-04-29T16:26:34

In this way:

$2x^2+6x-3=0rArr2(x^2+3x)-3=0rArr$

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

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

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

$2(x+3/2)^2=(9+6)/2rArr(x+3/2)^2=15/4rArr$

$x+3/2=+-sqrt15/2rArrx=-3/2+-sqrt15/2$ .

How do you solve 2x^2+6x-3=02x^2+6x-3=0 by completing the square?