What are the solutions to the equation roots x^2+x+3 ?

The solutions to the equation roots x^2+x+3 are x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2

Find the zeros of roots x^2+x+3

The zeros of roots x^2+x+3 are x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2

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roots

x^{2} + x + 3

x = - \frac{1}{2} + i \frac{11}{2}, x = - \frac{1}{2} - i \frac{11}{2}

Solution steps

x^{2} + x + 3

The roots are the intercepts with the x-axis

(y = 0)

x^{2} + x + 3 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 3 : 11 i

x_{1, 2} = \frac{- 1 \pm 11 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 11 i}{2 \cdot 1}, x_{2} = \frac{- 1 - 11 i}{2 \cdot 1}

x = \frac{- 1 + 11 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{11}{2}

x = \frac{- 1 - 11 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{11}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2} + i \frac{11}{2}, x = - \frac{1}{2} - i \frac{11}{2}

x^{2} + x + 6

4 w^{2} - 19 w + 21 = 2

y, x^{2} + y^{2} - 4 x + 6 y + 13 = 0

2^{2} + 21^{2} = x^{2}

9 x^{2} - 6 x - 7 = 0

What are the solutions to the equation roots x^2+x+3 ?
The solutions to the equation roots x^2+x+3 are x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2
Find the zeros of roots x^2+x+3
The zeros of roots x^2+x+3 are x=-1/2+i(sqrt(11))/2 ,x=-1/2-i(sqrt(11))/2