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

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

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

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

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y^{2} + 2 y = 0

y = 0, y = - 2

Solution steps

y^{2} + 2 y = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 0, y = - 2

2 4^{2} + 3 2^{2} = x^{2}

\frac{1}{2} \cdot \frac{1}{2}

\frac{3}{4} \cdot \frac{4}{3}

\frac{( \frac{1}{x + h} - \frac{1}{x} )}{h}

\frac{1}{3 ^{3}}

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