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

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

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

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

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- 2 = x^{2} + 2 x

x = - 1 + i, x = - 1 - i

Solution steps

- 2 = x^{2} + 2 x

Switch sides

x^{2} + 2 x = - 2

Move

2

to the left side

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 \cdot 1 \cdot 2 : 2 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + i, x = - 1 - i

v, F = \frac{z M v ^{2}}{r}

k^{2} - 14 k - 28 = 0

5 y + 3 y^{2} = 28

2 a^{2} - 13 a + 6 = 0

10 x^{2} - 9 x - 9 = 0

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