What are the solutions to the equation 3x^2+14x+1=0 ?

The solutions to the equation 3x^2+14x+1=0 are x=(-7+sqrt(46))/3 ,x=-(7+sqrt(46))/3

Find the zeros of 3x^2+14x+1=0

The zeros of 3x^2+14x+1=0 are x=(-7+sqrt(46))/3 ,x=-(7+sqrt(46))/3

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3 x^{2} + 14 x + 1 = 0

x = \frac{- 7 + 46}{3}, x = - \frac{7 + 46}{3}

Decimal Notation

x = - 0.07255 \dots, x = - 4.59410 \dots

Solution steps

3 x^{2} + 14 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

1 4^{2} - 4 \cdot 3 \cdot 1 = 246

x_{1, 2} = \frac{- 14 \pm 2 46}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 14 + 2 46}{2 \cdot 3}, x_{2} = \frac{- 14 - 2 46}{2 \cdot 3}

x = \frac{- 14 + 2 46}{2 \cdot 3} : \frac{- 7 + 46}{3}

x = \frac{- 14 - 2 46}{2 \cdot 3} : - \frac{7 + 46}{3}

The solutions to the quadratic equation are:

x = \frac{- 7 + 46}{3}, x = - \frac{7 + 46}{3}

\frac{1}{3} ∣ 3 x - 1 ∣ - 3 \leq 5

\frac{1}{11} x^{2} - 11 = 0

\frac{2 3}{6}

3 x^{4} - 15 n^{3} + 12 n^{2}

∣ x - 7 ∣ \geq 8

What are the solutions to the equation 3x^2+14x+1=0 ?
The solutions to the equation 3x^2+14x+1=0 are x=(-7+sqrt(46))/3 ,x=-(7+sqrt(46))/3
Find the zeros of 3x^2+14x+1=0
The zeros of 3x^2+14x+1=0 are x=(-7+sqrt(46))/3 ,x=-(7+sqrt(46))/3