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 \cdot 2 \cdot 0}{2 \cdot 2}

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

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

Separate the solutions

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

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

x = \frac{- 4 - 4}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = 0, x = - 2

8 - 2

1 1^{2}

- 2 x = 16

6 (x + 5) + 2 = 8 (2 x - 5) + 22

8 < - 2 (x - 3)

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