What are the solutions to the equation x^2+2x=-2x-5 ?

The solutions to the equation x^2+2x=-2x-5 are x=-2+i,x=-2-i

Find the zeros of x^2+2x=-2x-5

The zeros of x^2+2x=-2x-5 are x=-2+i,x=-2-i

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

x = - 2 + i, x = - 2 - i

Solution steps

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

Move

5

to the left side

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

Move

2 x

to the left side

x^{2} + 4 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 5 : 2 i

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

Separate the solutions

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

x = \frac{- 4 + 2 i}{2 \cdot 1} : - 2 + i

x = \frac{- 4 - 2 i}{2 \cdot 1} : - 2 - i

The solutions to the quadratic equation are:

x = - 2 + i, x = - 2 - i

3 y - y^{2} - 4 = 0

s^{2} + 14 s + 2 = 0

6 x^{2} + x - 5 = 0

- x - x^{2} = 0

x^{2} + 4 x = 12

What are the solutions to the equation x^2+2x=-2x-5 ?
The solutions to the equation x^2+2x=-2x-5 are x=-2+i,x=-2-i
Find the zeros of x^2+2x=-2x-5
The zeros of x^2+2x=-2x-5 are x=-2+i,x=-2-i