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

The solutions to the equation 2s^2+s+3=0 are s=-1/4+i(sqrt(23))/4 ,s=-1/4-i(sqrt(23))/4

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

The zeros of 2s^2+s+3=0 are s=-1/4+i(sqrt(23))/4 ,s=-1/4-i(sqrt(23))/4

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2 s^{2} + s + 3 = 0

s = - \frac{1}{4} + i \frac{23}{4}, s = - \frac{1}{4} - i \frac{23}{4}

Solution steps

2 s^{2} + s + 3 = 0

Solve with the quadratic formula

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

Simplify

1^{2} - 4 \cdot 2 \cdot 3 : 23 i

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

Separate the solutions

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

s = \frac{- 1 + 23 i}{2 \cdot 2} : - \frac{1}{4} + i \frac{23}{4}

s = \frac{- 1 - 23 i}{2 \cdot 2} : - \frac{1}{4} - i \frac{23}{4}

The solutions to the quadratic equation are:

s = - \frac{1}{4} + i \frac{23}{4}, s = - \frac{1}{4} - i \frac{23}{4}

3 - x^{2} \leq - 2 x + 4

x^{2} - \frac{5}{6} x + \frac{1}{6} = 0

1058

m^{2} + 2 m = 8

- 1 \cdot - 1

What are the solutions to the equation 2s^2+s+3=0 ?
The solutions to the equation 2s^2+s+3=0 are s=-1/4+i(sqrt(23))/4 ,s=-1/4-i(sqrt(23))/4
Find the zeros of 2s^2+s+3=0
The zeros of 2s^2+s+3=0 are s=-1/4+i(sqrt(23))/4 ,s=-1/4-i(sqrt(23))/4