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

The solutions to the equation -3s^2+s+1=0 are s=-(-1+sqrt(13))/6 ,s=(1+sqrt(13))/6

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

The zeros of -3s^2+s+1=0 are s=-(-1+sqrt(13))/6 ,s=(1+sqrt(13))/6

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

s = - \frac{- 1 + 13}{6}, s = \frac{1 + 13}{6}

Decimal Notation

s = - 0.43425 \dots, s = 0.76759 \dots

Solution steps

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

Solve with the quadratic formula

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

1^{2} - 4 (- 3) \cdot 1 = 13

s_{1, 2} = \frac{- 1 \pm 13}{2 ( - 3 )}

Separate the solutions

s_{1} = \frac{- 1 + 13}{2 ( - 3 )}, s_{2} = \frac{- 1 - 13}{2 ( - 3 )}

s = \frac{- 1 + 13}{2 ( - 3 )} : - \frac{- 1 + 13}{6}

s = \frac{- 1 - 13}{2 ( - 3 )} : \frac{1 + 13}{6}

The solutions to the quadratic equation are:

s = - \frac{- 1 + 13}{6}, s = \frac{1 + 13}{6}

2 x^{2} - 4 = 0

4^{3}

230 + 20 x - 0.5 x^{2} = 0

13 x^{2} - 5 x \leq 0

7 b = 49

What are the solutions to the equation -3s^2+s+1=0 ?
The solutions to the equation -3s^2+s+1=0 are s=-(-1+sqrt(13))/6 ,s=(1+sqrt(13))/6
Find the zeros of -3s^2+s+1=0
The zeros of -3s^2+s+1=0 are s=-(-1+sqrt(13))/6 ,s=(1+sqrt(13))/6