How do you solve $2 y^{3} + 7 y^{2} - 30 y = 0$ $2y^3+7y^2-30y=0$ ?

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Answer 1 · 2018-03-24T13:06:17

$y_{1} = - 6$ $y_1=-6$ , $y_{2} = 0$ $y_2=0$ and $y_{3} = \frac{5}{2}$ $y_3=5/2$

$2 y^{3} + 7 y^{2} - 30 y = 0$ $2y^3+7y^2-30y=0$

$y \cdot (2 y^{2} + 7 y - 30) = 0$ $y*(2y^2+7y-30)=0$

$y \cdot (2 y^{2} + 12 y - 5 y - 30) = 0$ $y*(2y^2+12y-5y-30)=0$

$y \cdot (y + 6) \cdot (2 y - 5) = 0$ $y*(y+6)*(2y-5)=0$

Thus, $y_{1} = - 6$ $y_1=-6$ , $y_{2} = 0$ $y_2=0$ and $y_{3} = \frac{5}{2}$ $y_3=5/2$

How do you solve 2y3+7y2−30y=02y^3+7y^2-30y=0?