What are the solutions to the equation s^2+3s+1=0 ?

The solutions to the equation s^2+3s+1=0 are s=(-3+sqrt(5))/2 ,s=(-3-sqrt(5))/2

Find the zeros of s^2+3s+1=0

The zeros of s^2+3s+1=0 are s=(-3+sqrt(5))/2 ,s=(-3-sqrt(5))/2

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s^{2} + 3 s + 1 = 0

s = \frac{- 3 + 5}{2}, s = \frac{- 3 - 5}{2}

Decimal Notation

s = - 0.38196 \dots, s = - 2.61803 \dots

Solution steps

s^{2} + 3 s + 1 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot 1 = 5

s_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 3 + 5}{2 \cdot 1}, s_{2} = \frac{- 3 - 5}{2 \cdot 1}

s = \frac{- 3 + 5}{2 \cdot 1} : \frac{- 3 + 5}{2}

s = \frac{- 3 - 5}{2 \cdot 1} : \frac{- 3 - 5}{2}

The solutions to the quadratic equation are:

s = \frac{- 3 + 5}{2}, s = \frac{- 3 - 5}{2}

(\frac{x}{3})^{4}

\frac{\frac{240}{289}}{\frac{161}{289}}

2 x - 1 \leq 7

9 x^{2} - 80 x - 9

\frac{a}{7} - 4 = \frac{a}{2} + 11

What are the solutions to the equation s^2+3s+1=0 ?
The solutions to the equation s^2+3s+1=0 are s=(-3+sqrt(5))/2 ,s=(-3-sqrt(5))/2
Find the zeros of s^2+3s+1=0
The zeros of s^2+3s+1=0 are s=(-3+sqrt(5))/2 ,s=(-3-sqrt(5))/2