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

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

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

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

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6 y^{2} - 12 y = 0

y = 2, y = 0

Solution steps

6 y^{2} - 12 y = 0

Solve with the quadratic formula

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

(- 12)^{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} : 2

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

The solutions to the quadratic equation are:

y = 2, y = 0

8 x^{6} - 27 y^{9}

\frac{2 x + 1}{x + 1} \leq 2

x^{2} - 6 x + 2 = 0

- \frac{5.98 E 16}{( 6.77 \cdot 1 0 ^{6} ) ^{3}}

y^{2} = 2 + y

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