What are the solutions to the equation roots s^2+s+6 ?

The solutions to the equation roots s^2+s+6 are s=-1/2+i(sqrt(23))/2 ,s=-1/2-i(sqrt(23))/2

Find the zeros of roots s^2+s+6

The zeros of roots s^2+s+6 are s=-1/2+i(sqrt(23))/2 ,s=-1/2-i(sqrt(23))/2

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roots

s^{2} + s + 6

s = - \frac{1}{2} + i \frac{23}{2}, s = - \frac{1}{2} - i \frac{23}{2}

Solution steps

s^{2} + s + 6

The roots are the intercepts with the x-axis

(y = 0)

s^{2} + s + 6 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 1 \cdot 6 : 23 i

s_{1, 2} = \frac{- 1 \pm 23 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 1 + 23 i}{2 \cdot 1}, s_{2} = \frac{- 1 - 23 i}{2 \cdot 1}

s = \frac{- 1 + 23 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{23}{2}

s = \frac{- 1 - 23 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{23}{2}

The solutions to the quadratic equation are:

s = - \frac{1}{2} + i \frac{23}{2}, s = - \frac{1}{2} - i \frac{23}{2}

3 t^{2} - 16 t + 28 = 0

x (2 x - 1) = 120

4 z^{2} - 100 = - 80

x^{2} - 5 x + \frac{5}{4}

- \frac{1}{5} x^{2} + 20 = 0

What are the solutions to the equation roots s^2+s+6 ?
The solutions to the equation roots s^2+s+6 are s=-1/2+i(sqrt(23))/2 ,s=-1/2-i(sqrt(23))/2
Find the zeros of roots s^2+s+6
The zeros of roots s^2+s+6 are s=-1/2+i(sqrt(23))/2 ,s=-1/2-i(sqrt(23))/2