What are the solutions to the equation 4y^2-8y=0 ?

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

Find the zeros of 4y^2-8y=0

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

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4 y^{2} - 8 y = 0

y = 2, y = 0

Solution steps

4 y^{2} - 8 y = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 8}{2 \cdot 4}, y_{2} = \frac{- ( - 8 ) - 8}{2 \cdot 4}

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

y = \frac{- ( - 8 ) - 8}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

y = 2, y = 0

4 - 9 x^{2} = 0

r^{2} + 13 r + 42 = 0

x^{2} - 18 x = - 56

x^{2} - 23 x + 3 = 0

a, \frac{d ^{3}}{d t ^{3}} ((u) u^{3}) = a^{2} t

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