find the zeros of the quadratic equation 6x^2-3-7x
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6X² - 7x - 3
6x² - 9x + 2x - 3.
3x ( 2x - 3 ) + 1 ( 2x - 3 ) .
( 2x - 3 ) ( 3x + 1 ) = 0
(2x - 3 ) = 0 or ( 3x + 1 ) = 0
x = 3/2 or x = -1/3
6X² - 7x - 3
6x² - 9x + 2x - 3.
3x ( 2x - 3 ) + 1 ( 2x - 3 ) .
( 2x - 3 ) ( 3x + 1 ) = 0
(2x - 3 ) = 0 or ( 3x + 1 ) = 0
x = 3/2 or x = -1/3
vinny375:
hii
and thanks
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