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

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

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

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

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2 x^{2} - 8 x = 0

x = 4, x = 0

Solution steps

2 x^{2} - 8 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = 0

\frac{12 x ^{- 5} y ^{3}}{3 ^{- 2} x ^{- 2} y ^{- 4}}

x^{2} > 0

2 x^{2} - 5 x + 1 < 0

21 x + 5 x^{2} = - 4

- - 13

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