What are the solutions to the equation solvefor x,x^2+2x+2=0 ?

The solutions to the equation solvefor x,x^2+2x+2=0 are x=-1+i,x=-1-i

Find the zeros of solvefor x,x^2+2x+2=0

The zeros of solvefor x,x^2+2x+2=0 are x=-1+i,x=-1-i

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

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

Solution steps

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

x^{2} - 8 = 2 x

9 (7 - 5 x)^{2} - 3 = 22

4 x^{2} = 24 x - 39

(20 - n) (20 + n) = 391

5 x^{2} - 60 x = 0

What are the solutions to the equation solvefor x,x^2+2x+2=0 ?
The solutions to the equation solvefor x,x^2+2x+2=0 are x=-1+i,x=-1-i
Find the zeros of solvefor x,x^2+2x+2=0
The zeros of solvefor x,x^2+2x+2=0 are x=-1+i,x=-1-i