What are the solutions to the equation s^2+s+3=0 ?

The solutions to the equation s^2+s+3=0 are s=-1/2+i(sqrt(11))/2 ,s=-1/2-i(sqrt(11))/2

Find the zeros of s^2+s+3=0

The zeros of s^2+s+3=0 are s=-1/2+i(sqrt(11))/2 ,s=-1/2-i(sqrt(11))/2

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s^{2} + s + 3 = 0

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

Solution steps

s^{2} + s + 3 = 0

Solve with the quadratic formula

s_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{1 0 ^{- 15}}{3.9 \cdot 1 0 ^{- 7}}

4 x + 5 \leq 8 x - 7

m^{9} - 27 n^{12}

x \cdot (100 - (\frac{1}{( 0.01 ) \cdot ( 1.01 ) ^{240}})) = 240000

2 x + 9 + 3 x + x = 6 x + 9

What are the solutions to the equation s^2+s+3=0 ?
The solutions to the equation s^2+s+3=0 are s=-1/2+i(sqrt(11))/2 ,s=-1/2-i(sqrt(11))/2
Find the zeros of s^2+s+3=0
The zeros of s^2+s+3=0 are s=-1/2+i(sqrt(11))/2 ,s=-1/2-i(sqrt(11))/2