What are the solutions to the equation 9y^2+12y=0 ?

The solutions to the equation 9y^2+12y=0 are y=0,y=-4/3

Find the zeros of 9y^2+12y=0

The zeros of 9y^2+12y=0 are y=0,y=-4/3

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9 y^{2} + 12 y = 0

y = 0, y = - \frac{4}{3}

Solution steps

9 y^{2} + 12 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 9 \cdot 0}{2 \cdot 9}

1 2^{2} - 4 \cdot 9 \cdot 0 = 12

y_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 9}

Separate the solutions

y_{1} = \frac{- 12 + 12}{2 \cdot 9}, y_{2} = \frac{- 12 - 12}{2 \cdot 9}

y = \frac{- 12 + 12}{2 \cdot 9} : 0

y = \frac{- 12 - 12}{2 \cdot 9} : - \frac{4}{3}

The solutions to the quadratic equation are:

y = 0, y = - \frac{4}{3}

3 x^{2} + 14 x = - 11

(11 - x) (6 + x) = 2 (x - 2)^{2}

6 x^{2} - 29 x - 534 = 7 x

5 - (x + 3) = 4 x^{2}

u^{2} - 40 u + 400 = 16

What are the solutions to the equation 9y^2+12y=0 ?
The solutions to the equation 9y^2+12y=0 are y=0,y=-4/3
Find the zeros of 9y^2+12y=0
The zeros of 9y^2+12y=0 are y=0,y=-4/3