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

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

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

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

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7 x^{2} + 4 x = 2

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

Move

2

to the left side

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}

2 x^{2} + x + 8 = 0

(x + 1) (x - 10) = 0

8 - \frac{3 x}{10} + \frac{2 x}{4} - \frac{5 x}{8} = - 9

4 x - \frac{2}{5} > \frac{3}{5}

19 - 2 (1 - x) < 13

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