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Solve this equation by completing the square. 3x2 - 4x = -2
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3x^2-4x=-2 3x^2=4x+2=0 (x )(x )=0
(3x )( x )=0
i know i got to (3x ) (x )=0 but i tried all the numbers i could think of in the spaces
use this formula [(-b)+- sqr(b^2-4ac)]/2a
b=-4 a=3 c=2
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i dont understand how to input my equation into the one you gave me. which parts are A,B,and C
[4 +- sqr(16-4*3*2)]/6 [4+-sqr(16-24)/]6 [4+-aqr(-8)]/6 2/3+-2/6*sqr(-2)
A: x^2 coefficient B: x coefficient C: the number
okay thank you a ton
the sqr of negative number irrational, then the two roots are a conjugate pair.
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