How do you solve $4x^2+6x=12$ by completing the square?

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Answer 1 · 2017-03-21T12:40:15

The solutions are $S={-2.637, 1.137}$

Before completing the square, we divide by $2$

$4x^2+6x=12$

$2x^2+3x=6$

$2(x^2+3/2x)=6$

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

Now, we complete the square

$(x^2+3/2x+9/16)=3+9/16$

$(x+3/4)^2=57/16$

$x+3/4=+-sqrt57/16$

$x=-3/4+-sqrt57/16$

$x=1/4(-3+-sqrt57)$

Therefore,

$x_1=1/4(-3-sqrt57)=-2.637$

$x_2=1/4(-3+sqrt57)=1.137$

How do you solve 4x^2+6x=124x^2+6x=12 by completing the square?