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

The solutions to the equation -7x^2+2x-1=0 are x= 1/7-i(sqrt(6))/7 ,x= 1/7+i(sqrt(6))/7

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

The zeros of -7x^2+2x-1=0 are x= 1/7-i(sqrt(6))/7 ,x= 1/7+i(sqrt(6))/7

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

x = \frac{1}{7} - i \frac{6}{7}, x = \frac{1}{7} + i \frac{6}{7}

Solution steps

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

Solve with the quadratic formula

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

Simplify

2^{2} - 4 (- 7) (- 1) : 26 i

x_{1, 2} = \frac{- 2 \pm 2 6 i}{2 ( - 7 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 6 i}{2 ( - 7 )}, x_{2} = \frac{- 2 - 2 6 i}{2 ( - 7 )}

x = \frac{- 2 + 2 6 i}{2 ( - 7 )} : \frac{1}{7} - i \frac{6}{7}

x = \frac{- 2 - 2 6 i}{2 ( - 7 )} : \frac{1}{7} + i \frac{6}{7}

The solutions to the quadratic equation are:

x = \frac{1}{7} - i \frac{6}{7}, x = \frac{1}{7} + i \frac{6}{7}

x^{2} = 47

- 3.35179 \cdot \frac{t ^{2}}{2} + 4.1 t = 0

4 x - 1 \leq 5 x \leq 3 (x + 2)

(x + 9) (x - 3) = (x + 1)^{2}

(- \frac{27}{125})^{\frac{- 2}{3}}

What are the solutions to the equation -7x^2+2x-1=0 ?
The solutions to the equation -7x^2+2x-1=0 are x= 1/7-i(sqrt(6))/7 ,x= 1/7+i(sqrt(6))/7
Find the zeros of -7x^2+2x-1=0
The zeros of -7x^2+2x-1=0 are x= 1/7-i(sqrt(6))/7 ,x= 1/7+i(sqrt(6))/7