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

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

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

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

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x^{2} + 2 x = 0

x = 0, x = - 2

Solution steps

x^{2} + 2 x = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot 0 = 2

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

Separate the solutions

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

x = \frac{- 2 + 2}{2 \cdot 1} : 0

x = \frac{- 2 - 2}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 0, x = - 2

2 x^{2} - 16 x + 38

m, 6 m^{2} - 8 m + 1 = 0

6 x^{2} - 5 x - 2 = - 6 x^{2}

4 (x + 7)^{2} = 13

- 4.905 t^{2} + 12 t + 80 = 0

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