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

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

Find the zeros of roots s^2+3s+1

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

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roots

s^{2} + 3 s + 1

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

Solution steps

s^{2} + 3 s + 1

The roots are the intercepts with the x-axis

(y = 0)

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}

2 x^{2} - 8 = 100

47 = 73 t - 5 t^{2}

- \frac{1}{2} x (2 x + 3) = 0

(x - 1)^{2} = (2 x - 2)

- 4 x^{2} + 32 x + 0 = 0

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