What are the solutions to the equation 3y^2+18y=0 ?

The solutions to the equation 3y^2+18y=0 are y=0,y=-6

Find the zeros of 3y^2+18y=0

The zeros of 3y^2+18y=0 are y=0,y=-6

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3 y^{2} + 18 y = 0

y = 0, y = - 6

Solution steps

3 y^{2} + 18 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

1 8^{2} - 4 \cdot 3 \cdot 0 = 18

y_{1, 2} = \frac{- 18 \pm 18}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 18 + 18}{2 \cdot 3}, y_{2} = \frac{- 18 - 18}{2 \cdot 3}

y = \frac{- 18 + 18}{2 \cdot 3} : 0

y = \frac{- 18 - 18}{2 \cdot 3} : - 6

The solutions to the quadratic equation are:

y = 0, y = - 6

\frac{x}{2} - 12 = - 6

4 a^{2} - 9 a + 5

21 (x - 3)^{2} + 7 (x - 3)

(\frac{7}{4})^{2} + 1

5 (x - 8) + 15 x = - 35

What are the solutions to the equation 3y^2+18y=0 ?
The solutions to the equation 3y^2+18y=0 are y=0,y=-6
Find the zeros of 3y^2+18y=0
The zeros of 3y^2+18y=0 are y=0,y=-6