What are the solutions to the equation 3s^2+10s+4=0 ?

The solutions to the equation 3s^2+10s+4=0 are s=(-5+sqrt(13))/3 ,s=-(5+sqrt(13))/3

Find the zeros of 3s^2+10s+4=0

The zeros of 3s^2+10s+4=0 are s=(-5+sqrt(13))/3 ,s=-(5+sqrt(13))/3

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3 s^{2} + 10 s + 4 = 0

s = \frac{- 5 + 13}{3}, s = - \frac{5 + 13}{3}

Decimal

s = - 0.46481 \dots, s = - 2.86851 \dots

Solution steps

3 s^{2} + 10 s + 4 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 4}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 \cdot 4 = 213

s_{1, 2} = \frac{- 10 \pm 2 13}{2 \cdot 3}

Separate the solutions

s_{1} = \frac{- 10 + 2 13}{2 \cdot 3}, s_{2} = \frac{- 10 - 2 13}{2 \cdot 3}

s = \frac{- 10 + 2 13}{2 \cdot 3} : \frac{- 5 + 13}{3}

s = \frac{- 10 - 2 13}{2 \cdot 3} : - \frac{5 + 13}{3}

The solutions to the quadratic equation are:

s = \frac{- 5 + 13}{3}, s = - \frac{5 + 13}{3}

6 x^{2} = 15 x

9 x^{2} + 18 x - 5 = 0

- 2 x^{2} + 3 x = 1

x^{2} - 2 x - 60 = 0

1 = a (2 x^{2} - x) + b (3 x^{2} + 1) + c (x^{2})

What are the solutions to the equation 3s^2+10s+4=0 ?
The solutions to the equation 3s^2+10s+4=0 are s=(-5+sqrt(13))/3 ,s=-(5+sqrt(13))/3
Find the zeros of 3s^2+10s+4=0
The zeros of 3s^2+10s+4=0 are s=(-5+sqrt(13))/3 ,s=-(5+sqrt(13))/3