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

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Answer 1 · 2016-11-10T14:03:29

$x=(3+-sqrt(15))/2$

$2x^2-3x-3=0$

divide by $2$ to make coefficient $x^2=1$

$x^2-(3x)/2-3/2=0$

$(x^2-(3x)/2)-3/2=0$

square half coefficient of $x$ then add and subtract

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

the first bracket is now a perfect square

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

$(x-3/2)^2-15/4=0$

now rearrange for $x$

$(x-3/2)^2=15/4$

$x-3/2=+-sqrt(15/4)$

$x-3/2=+-sqrt(15)/2$

$x=3/2+-sqrt(15)/2$

$x=(3+-sqrt(15))/2$

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