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

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

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

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

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

x = \frac{1 + 2 2}{7}, x = \frac{1 - 2 2}{7}

Decimal Notation

x = 0.54691 \dots, x = - 0.26120 \dots

Solution steps

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

Switch sides

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 7 (- 1) = 42

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

Separate the solutions

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

x = \frac{- ( - 2 ) + 4 2}{2 \cdot 7} : \frac{1 + 2 2}{7}

x = \frac{- ( - 2 ) - 4 2}{2 \cdot 7} : \frac{1 - 2 2}{7}

The solutions to the quadratic equation are:

x = \frac{1 + 2 2}{7}, x = \frac{1 - 2 2}{7}

a, a^{2} + 2 ab - b^{2} = 0

t^{2} - 4.5 t + 3.5 = 0

6 x^{2} + 3 x = 0

3 p (2 p + 5) = 0

x^{2} + 2 b x = a^{2} - b^{2}

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