What are the solutions to the equation 6y^2+12y=0 ?

The solutions to the equation 6y^2+12y=0 are y=0,y=-2

Find the zeros of 6y^2+12y=0

The zeros of 6y^2+12y=0 are y=0,y=-2

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6 y^{2} + 12 y = 0

y = 0, y = - 2

Solution steps

6 y^{2} + 12 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

1 2^{2} - 4 \cdot 6 \cdot 0 = 12

y_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- 12 + 12}{2 \cdot 6}, y_{2} = \frac{- 12 - 12}{2 \cdot 6}

y = \frac{- 12 + 12}{2 \cdot 6} : 0

y = \frac{- 12 - 12}{2 \cdot 6} : - 2

The solutions to the quadratic equation are:

y = 0, y = - 2

\frac{\frac{π ( 50 ) ^{2}}{4} 200}{600} (- 0.162)

- \frac{2}{5} y = \frac{4}{15}

\frac{2 x + x}{3} = x - 5

5 x + \frac{x}{3} = 32

- 3 x^{2} - 12 x - 9

What are the solutions to the equation 6y^2+12y=0 ?
The solutions to the equation 6y^2+12y=0 are y=0,y=-2
Find the zeros of 6y^2+12y=0
The zeros of 6y^2+12y=0 are y=0,y=-2