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

The solutions to the equation 12y-12y^2+6y=0 are y=0,y= 3/2

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

The zeros of 12y-12y^2+6y=0 are y=0,y= 3/2

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12 y - 12 y^{2} + 6 y = 0

y = 0, y = \frac{3}{2}

Decimal Notation

y = 0, y = 1.5

Solution steps

12 y - 12 y^{2} + 6 y = 0

Refine

- 12 y^{2} + 18 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 12 ) \cdot 0}{2 ( - 12 )}

1 8^{2} - 4 (- 12) \cdot 0 = 18

y_{1, 2} = \frac{- 18 \pm 18}{2 ( - 12 )}

Separate the solutions

y_{1} = \frac{- 18 + 18}{2 ( - 12 )}, y_{2} = \frac{- 18 - 18}{2 ( - 12 )}

y = \frac{- 18 + 18}{2 ( - 12 )} : 0

y = \frac{- 18 - 18}{2 ( - 12 )} : \frac{3}{2}

The solutions to the quadratic equation are:

y = 0, y = \frac{3}{2}

3 x = 2 x - (6 - 3 x)

∣ x - 4 ∣ = ∣ 3 x + 9 ∣

4 \cdot (2 x - 4) = 6 \cdot (x - 1)

y, x^{3} - 3 x^{2} y + 2 x y^{2} = 12

\frac{2 + 2 i}{i}

What are the solutions to the equation 12y-12y^2+6y=0 ?
The solutions to the equation 12y-12y^2+6y=0 are y=0,y= 3/2
Find the zeros of 12y-12y^2+6y=0
The zeros of 12y-12y^2+6y=0 are y=0,y= 3/2