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

The solutions to the equation (x)=x^2+2x+1 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2

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

The zeros of (x)=x^2+2x+1 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2

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

x = - \frac{1}{2} + i \frac{3}{2}, x = - \frac{1}{2} - i \frac{3}{2}

Solution steps

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

Refine

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

Switch sides

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

Move

x

to the left side

x^{2} + x + 1 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{3}{2}, x = - \frac{1}{2} - i \frac{3}{2}

D, (D^{2} - 4 D + 5) 3 (D - 3) 4 y = 0

q^{2} + 3 q - 4 = 0

- 0.0013 x^{2} + x + 13 = 0

X^{2} + 10 X + 21 = 0

- 16 x^{2} + 272 x + 204 = 0

What are the solutions to the equation (x)=x^2+2x+1 ?
The solutions to the equation (x)=x^2+2x+1 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2
Find the zeros of (x)=x^2+2x+1
The zeros of (x)=x^2+2x+1 are x=-1/2+i(sqrt(3))/2 ,x=-1/2-i(sqrt(3))/2