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

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

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

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

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x^{2} + (x + 1)^{2} + (x + 2)^{2} = - 1

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

Solution steps

x^{2} + (x + 1)^{2} + (x + 2)^{2} = - 1

Expand

x^{2} + (x + 1)^{2} + (x + 2)^{2} : 3 x^{2} + 6 x + 5

3 x^{2} + 6 x + 5 = - 1

Move

1

to the left side

3 x^{2} + 6 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

Simplify

6^{2} - 4 \cdot 3 \cdot 6 : 6 i

x_{1, 2} = \frac{- 6 \pm 6 i}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 6 i}{2 \cdot 3}, x_{2} = \frac{- 6 - 6 i}{2 \cdot 3}

x = \frac{- 6 + 6 i}{2 \cdot 3} : - 1 + i

x = \frac{- 6 - 6 i}{2 \cdot 3} : - 1 - i

The solutions to the quadratic equation are:

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

x^{2} - 12 x + 3 = 0

4 x^{2} - 28 x + 53 = 0

x^{2} + 4 x - 2

x^{2} + 5 + 6 = 0

s^{2} + 2 s + 3

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