Asked by Mohamed

3y^2-5y+2=0

6 years ago

Answered by Reiny

3y^2-5y+2=0
(3y + 1)(y - 2) = 0

you know what to do from here.

6 years ago

Answered by henry2,

3y^2 - 5y + 2 = 0.
A*C = 3*2 = 6 = (-2)*(-3). sum = -2 + (-3) = -5 = B.
3y^2 + (-2y-3y) + 2 = 0,
(3y^2 - 2y) + (-3y + 2) = 0,
y(3y - 2) - (3y - 2) = 0.
(3y-2)(y - 1) = 0,
3y-2 = 0, Y = 2/3.
y-1 = 0, Y = 1.

6 years ago

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