What are the solutions to the equation 2y^2-4y=0 ?

The solutions to the equation 2y^2-4y=0 are y=2,y=0

Find the zeros of 2y^2-4y=0

The zeros of 2y^2-4y=0 are y=2,y=0

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2 y^{2} - 4 y = 0

y = 2, y = 0

Solution steps

2 y^{2} - 4 y = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 \cdot 0 = 4

y_{1, 2} = \frac{- ( - 4 ) \pm 4}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 4}{2 \cdot 2}, y_{2} = \frac{- ( - 4 ) - 4}{2 \cdot 2}

y = \frac{- ( - 4 ) + 4}{2 \cdot 2} : 2

y = \frac{- ( - 4 ) - 4}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

y = 2, y = 0

2 j^{2} + 4 j + 2 = 0

- 3 x^{2} - 4 x + 175 = 0

x, x^{2} - 4 x = y^{2} - 4 y

x^{2} + 5 x - 36 = 0

2 x - 5 = - x^{2}

What are the solutions to the equation 2y^2-4y=0 ?
The solutions to the equation 2y^2-4y=0 are y=2,y=0
Find the zeros of 2y^2-4y=0
The zeros of 2y^2-4y=0 are y=2,y=0