How do you solve $3 y^{2} + 2 y - 1 = 0$ $3y^2+2y-1=0$ ?

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How do you solve $3 y^{2} + 2 y - 1 = 0$ $3y^2+2y-1=0$ ?

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Answer 1 · 2018-03-30T15:01:38

$y = - 1, \frac{1}{3}$ $y=-1, 1/3$

Explanation:

$3 y^{2} + 3 y - y - 1 = 0$ $3y^2+3y-y-1=0$ find two integers that ADD to 2, MULTIPLY to -3 $(3 \cdot - 1)$ $(3*-1)$
$3 y (y + 1) - (y + 1)$ $3y(y+1)-(y+1)$ factor out common factor
$(3 y - 1) (y + 1)$ $(3y-1)(y+1)$
$3 y - 1 = 0$ $3y-1 = 0$ OR $y + 1 = 0$ $y+1=0$
$y = \frac{1}{3}$ $y=1/3$ OR $y = - 1$ $y=-1$

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How do you solve $3 y^{2} + 2 y - 1 = 0$ $3y^2+2y-1=0$ ?

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