What are the solutions to the equation 3s^2+18s+14=0 ?

The solutions to the equation 3s^2+18s+14=0 are s=(-9+sqrt(39))/3 ,s=-(9+sqrt(39))/3

Find the zeros of 3s^2+18s+14=0

The zeros of 3s^2+18s+14=0 are s=(-9+sqrt(39))/3 ,s=-(9+sqrt(39))/3

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3 s^{2} + 18 s + 14 = 0

s = \frac{- 9 + 39}{3}, s = - \frac{9 + 39}{3}

Decimal Notation

s = - 0.91833 \dots, s = - 5.08166 \dots

Solution steps

3 s^{2} + 18 s + 14 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 3 \cdot 14}{2 \cdot 3}

1 8^{2} - 4 \cdot 3 \cdot 14 = 239

s_{1, 2} = \frac{- 18 \pm 2 39}{2 \cdot 3}

Separate the solutions

s_{1} = \frac{- 18 + 2 39}{2 \cdot 3}, s_{2} = \frac{- 18 - 2 39}{2 \cdot 3}

s = \frac{- 18 + 2 39}{2 \cdot 3} : \frac{- 9 + 39}{3}

s = \frac{- 18 - 2 39}{2 \cdot 3} : - \frac{9 + 39}{3}

The solutions to the quadratic equation are:

s = \frac{- 9 + 39}{3}, s = - \frac{9 + 39}{3}

x, z = x^{2} y + 1

A, A BX A - 1 B - 1 = I + A

(x + 3)^{2} - (x - 3)^{2} = x^{2} + 32

y, x^{3} + x y^{2} - y^{2} = 0

- 5 q^{2} + 9 q + 13 = - 6 q^{2}

What are the solutions to the equation 3s^2+18s+14=0 ?
The solutions to the equation 3s^2+18s+14=0 are s=(-9+sqrt(39))/3 ,s=-(9+sqrt(39))/3
Find the zeros of 3s^2+18s+14=0
The zeros of 3s^2+18s+14=0 are s=(-9+sqrt(39))/3 ,s=-(9+sqrt(39))/3