What are the solutions to the equation 3x^2+19x-1=0 ?

The solutions to the equation 3x^2+19x-1=0 are x=(-19+sqrt(373))/6 ,x=(-19-sqrt(373))/6

Find the zeros of 3x^2+19x-1=0

The zeros of 3x^2+19x-1=0 are x=(-19+sqrt(373))/6 ,x=(-19-sqrt(373))/6

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3 x^{2} + 19 x - 1 = 0

x = \frac{- 19 + 373}{6}, x = \frac{- 19 - 373}{6}

Decimal Notation

x = 0.05220 \dots, x = - 6.38553 \dots

Solution steps

3 x^{2} + 19 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 3 ( - 1 )}{2 \cdot 3}

1 9^{2} - 4 \cdot 3 (- 1) = 373

x_{1, 2} = \frac{- 19 \pm 373}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 19 + 373}{2 \cdot 3}, x_{2} = \frac{- 19 - 373}{2 \cdot 3}

x = \frac{- 19 + 373}{2 \cdot 3} : \frac{- 19 + 373}{6}

x = \frac{- 19 - 373}{2 \cdot 3} : \frac{- 19 - 373}{6}

The solutions to the quadratic equation are:

x = \frac{- 19 + 373}{6}, x = \frac{- 19 - 373}{6}

4 x^{2} + 4 = - 8 x

y, x = y^{2} + 4 z^{2} - 1

x, x y (x + 2) (y - 4) = 3

x^{2} - 8 x - 6 = 4

- 3 x^{2} + 6 x + 21

What are the solutions to the equation 3x^2+19x-1=0 ?
The solutions to the equation 3x^2+19x-1=0 are x=(-19+sqrt(373))/6 ,x=(-19-sqrt(373))/6
Find the zeros of 3x^2+19x-1=0
The zeros of 3x^2+19x-1=0 are x=(-19+sqrt(373))/6 ,x=(-19-sqrt(373))/6