What are the solutions to the equation -1=x^2+2x ?

The solutions to the equation -1=x^2+2x are x=-1

Find the zeros of -1=x^2+2x

The zeros of -1=x^2+2x are x=-1

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- 1 = x^{2} + 2 x

x = - 1

Solution steps

- 1 = x^{2} + 2 x

Switch sides

x^{2} + 2 x = - 1

Move

1

to the left side

x^{2} + 2 x + 1 = 0

Solve with the quadratic formula

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

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

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

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

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

x = - 1

The solution to the quadratic equation is:

x = - 1

- x^{2} + 18 x - 80 = 0

4 8^{2} + x^{2} = 7 3^{2}

4 x^{2} + x + 5 = 9 x + 2

- 0.025 x^{2} + 16.24 x - 2445.2 = 0

11 = 3 x^{2} + 2 x

What are the solutions to the equation -1=x^2+2x ?
The solutions to the equation -1=x^2+2x are x=-1
Find the zeros of -1=x^2+2x
The zeros of -1=x^2+2x are x=-1