help please2y^2 - 5y + 2 = 0?

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2007-10-10 13:07:58 · 5 answers · asked by Onika D 1 in Science & Mathematics ➔ Mathematics

(2y - 1)(y - 2)

y= 1/2 and 2

2007-10-10 13:11:53 · answer #1 · answered by a.pasternak 2 · 0⤊ 0⤋

Factor and solve for y:

(2y - 1) (y - 2) = 0
2y - 1 = 0 and y - 2 = 0
2y = 1 and y = 2
y = 1/2 and 2

:)

2007-10-10 20:12:12 · answer #2 · answered by Alison 2 · 1⤊ 0⤋

2y^2 - 5y + 2 = 0
2y^2 - 4y - y + 2 = 0
2y(y - 2) - 1(y - 2) = 0
(2y - 1)(y - 2) = 0
y = 1/2 and y = 2

2007-10-10 20:13:13 · answer #3 · answered by sfroggy5 6 · 0⤊ 0⤋

Factor

( 2y - 1) ( y - 2)

2 and .5

2007-10-10 20:12:56 · answer #4 · answered by Anonymous · 0⤊ 0⤋

=(2y-1)(y-2)
so y = 1/2 and 2

2007-10-10 20:12:48 · answer #5 · answered by ironduke8159 7 · 0⤊ 0⤋