Posted by **Anonymous** on Tuesday, April 26, 2011 at 2:54pm.

how do i solve this completing the square problem: 3x^2+12x+10=0

- trig -
**MathMate**, Tuesday, April 26, 2011 at 3:31pm
3x^2+12x+10=0

3(x^2+4x+10/3)=0

(x+2)²-4+10/3=0

(x+2)²-(√(2/3))²=0

(x+2+√(2/3))(x+2-√(2/3))=0

therefore

x=-2-√(2/3) or -2+√(2/3)

- trig -
**bobpursley**, Tuesday, April 26, 2011 at 3:34pm
divide by 3

x^2+4x+10/3=0

x^2+4x + ???= -10/3 + ???

well, look at the second term. take one half of it, and square it, add to both sides.

x^2+4x+4=-10/3+4= 10/3+12/3

(x+2)^2= 2/3

take the square root of each side.

x+2= +-sqrt 2/3

x=-2+-sqrt 2/3

check that.

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