How do you solve the equation: $- 32 y^{2} - 24 y = 0$ $-32y^2-24y=0$ ?

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Answer 1 · 2015-06-28T07:27:10

The solutions are:
$y = 0, y = - \frac{3}{4}$ $color(blue)(y=0, y=-3/4$

$- 32 y^{2} - 24 y = 0$ $-32y^2-24y = 0$

$- 8 y$ $-8y$ is common to both terms:

$= - 8 y (4 y + 3) = 0$ $=-8y(4y + 3) = 0$

Equating the terms with zero to obtain the solution:

How do you solve the equation: −32y2−24y=0-32y^2-24y=0?