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

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

Find the zeros of 0=-2x^2+4x

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

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

x = 0, x = 2

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 4 + 4}{2 ( - 2 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 2

u^{2} - 18 u

9 - y^{2} = 0

(x - 2)^{2} + 4 (x - 2) = 0

2 x^{2} + x - 15 = 0

3 (x - 6)^{2} + 4 = 0

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