What are the solutions to the equation 3s^2+10s+6=0 ?

The solutions to the equation 3s^2+10s+6=0 are s=(-5+sqrt(7))/3 ,s=-(5+sqrt(7))/3

Find the zeros of 3s^2+10s+6=0

The zeros of 3s^2+10s+6=0 are s=(-5+sqrt(7))/3 ,s=-(5+sqrt(7))/3

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3 s^{2} + 10 s + 6 = 0

s = \frac{- 5 + 7}{3}, s = - \frac{5 + 7}{3}

Decimal Notation

s = - 0.78474 \dots, s = - 2.54858 \dots

Solution steps

3 s^{2} + 10 s + 6 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 \cdot 6 = 27

s_{1, 2} = \frac{- 10 \pm 2 7}{2 \cdot 3}

Separate the solutions

s_{1} = \frac{- 10 + 2 7}{2 \cdot 3}, s_{2} = \frac{- 10 - 2 7}{2 \cdot 3}

s = \frac{- 10 + 2 7}{2 \cdot 3} : \frac{- 5 + 7}{3}

s = \frac{- 10 - 2 7}{2 \cdot 3} : - \frac{5 + 7}{3}

The solutions to the quadratic equation are:

s = \frac{- 5 + 7}{3}, s = - \frac{5 + 7}{3}

(\frac{1}{32})^{\frac{3}{5}}

- \frac{1}{6} < x \leq 1.8

36 x^{2} - x = 0

(x - 4)^{2} = 36

\frac{1}{3} m < - 2

What are the solutions to the equation 3s^2+10s+6=0 ?
The solutions to the equation 3s^2+10s+6=0 are s=(-5+sqrt(7))/3 ,s=-(5+sqrt(7))/3
Find the zeros of 3s^2+10s+6=0
The zeros of 3s^2+10s+6=0 are s=(-5+sqrt(7))/3 ,s=-(5+sqrt(7))/3