What are the solutions to the equation 2y^2-8y=0 ?

The solutions to the equation 2y^2-8y=0 are y=4,y=0

Find the zeros of 2y^2-8y=0

The zeros of 2y^2-8y=0 are y=4,y=0

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2 y^{2} - 8 y = 0

y = 4, y = 0

Solution steps

2 y^{2} - 8 y = 0

Solve with the quadratic formula

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

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

y_{1, 2} = \frac{- ( - 8 ) \pm 8}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 8}{2 \cdot 2}, y_{2} = \frac{- ( - 8 ) - 8}{2 \cdot 2}

y = \frac{- ( - 8 ) + 8}{2 \cdot 2} : 4

y = \frac{- ( - 8 ) - 8}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

y = 4, y = 0

2 x^{2} + 9 x - 4 = 0

5 x^{2} + 2 = x

x^{2} + 8 x - 3 = 0

- x^{2} - 2 x + 26 = 0

24 = x^{2}

What are the solutions to the equation 2y^2-8y=0 ?
The solutions to the equation 2y^2-8y=0 are y=4,y=0
Find the zeros of 2y^2-8y=0
The zeros of 2y^2-8y=0 are y=4,y=0