How do you solve $3x^2 + 4x + 3 = 0$ by completing the square?

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Answer 1 · 2017-09-10T04:52:08

$x= +-(isqrt(-5))/3-2/3$

Given -

$3x^2+4x+3=0$
Take the constant term to right

$3x^2+4x=-3$

Divide each term on both sides by the coefficient of $x^2$

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

$x^2+4/3 x=-1$

Divide the coefficient $x$ , square it and add it to both sides

$x^2+2/3 x+4/9=-1+4/9=(-9+4)/9=-5/9$

$(x+2/3)^2=-5/9$
Take square root on both sides

$(x+2/3)=sqrt(-5/9)=sqrt(-5)/3$

$x= +-(isqrt(-5))/3-2/3$

How do you solve 3x^2 + 4x + 3 = 03x^2 + 4x + 3 = 0 by completing the square?