How do you solve $3x^2 + 10x - 2 = 0$ by completing the square?

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Answer 1 · 2017-10-07T06:56:55

$x=sqrt31/9-5/3$
$x=-sqrt31/9-5/3$

Given -

$3x^2+10x-2=0$

Take the constant term to right-hand side

$3x^2+10x=2$

Divide both sides by 3

$(3x^2)/3+(10x)/3=2/3$

$x^2+10/3x=2/3$

Divide the coefficient of $x$ by2; square it and add it to both sides

$x^2+10/3x+100/36=2/3+100/36=(24+100)/36=124/36=31/9$
$(x+10/6)^2=31/9$

$x+5/3=+-sqrt(31/9)= +-sqrt31/3$

$x=sqrt31/9-5/3$
$x=-sqrt31/9-5/3$

How do you solve 3x^2 + 10x - 2 = 0 3x^2 + 10x - 2 = 0 by completing the square?