What are the solutions to the equation 7x^2+4x-2=0 ?

The solutions to the equation 7x^2+4x-2=0 are x=(-2+3sqrt(2))/7 ,x=-(2+3sqrt(2))/7

Find the zeros of 7x^2+4x-2=0

The zeros of 7x^2+4x-2=0 are x=(-2+3sqrt(2))/7 ,x=-(2+3sqrt(2))/7

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7 x^{2} + 4 x - 2 = 0

x = \frac{- 2 + 3 2}{7}, x = - \frac{2 + 3 2}{7}

Decimal Notation

x = 0.32037 \dots, x = - 0.89180 \dots

Solution steps

7 x^{2} + 4 x - 2 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 7 (- 2) = 62

x_{1, 2} = \frac{- 4 \pm 6 2}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 4 + 6 2}{2 \cdot 7}, x_{2} = \frac{- 4 - 6 2}{2 \cdot 7}

x = \frac{- 4 + 6 2}{2 \cdot 7} : \frac{- 2 + 3 2}{7}

x = \frac{- 4 - 6 2}{2 \cdot 7} : - \frac{2 + 3 2}{7}

The solutions to the quadratic equation are:

x = \frac{- 2 + 3 2}{7}, x = - \frac{2 + 3 2}{7}

- (\frac{3}{2} + \frac{5}{2} i) + (\frac{5}{3} + \frac{11}{3} i)

- 2 x^{3} - 6 x^{2} + 56 x

x^{2} + 16 + 63

4 x^{2} - 4 x y - 3 y^{2}

- 3 < x < 4

What are the solutions to the equation 7x^2+4x-2=0 ?
The solutions to the equation 7x^2+4x-2=0 are x=(-2+3sqrt(2))/7 ,x=-(2+3sqrt(2))/7
Find the zeros of 7x^2+4x-2=0
The zeros of 7x^2+4x-2=0 are x=(-2+3sqrt(2))/7 ,x=-(2+3sqrt(2))/7