What are the solutions to the equation 7x^2+14x-1=0 ?

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

Find the zeros of 7x^2+14x-1=0

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

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7 x^{2} + 14 x - 1 = 0

x = \frac{- 7 + 2 14}{7}, x = - \frac{7 + 2 14}{7}

Decimal Notation

x = 0.06904 \dots, x = - 2.06904 \dots

Solution steps

7 x^{2} + 14 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 7 ( - 1 )}{2 \cdot 7}

1 4^{2} - 4 \cdot 7 (- 1) = 414

x_{1, 2} = \frac{- 14 \pm 4 14}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 14 + 4 14}{2 \cdot 7}, x_{2} = \frac{- 14 - 4 14}{2 \cdot 7}

x = \frac{- 14 + 4 14}{2 \cdot 7} : \frac{- 7 + 2 14}{7}

x = \frac{- 14 - 4 14}{2 \cdot 7} : - \frac{7 + 2 14}{7}

The solutions to the quadratic equation are:

x = \frac{- 7 + 2 14}{7}, x = - \frac{7 + 2 14}{7}

5 x^{2} + 30 x + 25 = 0

\frac{1 0 ^{- 2} x ^{- 4}}{y ^{0}}

\frac{1}{100} (x - 2)^{2} \leq 0

1 6^{\frac{1}{4}}

- 8 x - 9 = - 1

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