What are the solutions to the equation 7x^2-3x+1=0 ?

The solutions to the equation 7x^2-3x+1=0 are x= 3/14+i(sqrt(19))/(14),x= 3/14-i(sqrt(19))/(14)

Find the zeros of 7x^2-3x+1=0

The zeros of 7x^2-3x+1=0 are x= 3/14+i(sqrt(19))/(14),x= 3/14-i(sqrt(19))/(14)

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7 x^{2} - 3 x + 1 = 0

x = \frac{3}{14} + i \frac{19}{14}, x = \frac{3}{14} - i \frac{19}{14}

Solution steps

7 x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

Simplify

(- 3)^{2} - 4 \cdot 7 \cdot 1 : 19 i

x_{1, 2} = \frac{- ( - 3 ) \pm 19 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 19 i}{2 \cdot 7}, x_{2} = \frac{- ( - 3 ) - 19 i}{2 \cdot 7}

x = \frac{- ( - 3 ) + 19 i}{2 \cdot 7} : \frac{3}{14} + i \frac{19}{14}

x = \frac{- ( - 3 ) - 19 i}{2 \cdot 7} : \frac{3}{14} - i \frac{19}{14}

The solutions to the quadratic equation are:

x = \frac{3}{14} + i \frac{19}{14}, x = \frac{3}{14} - i \frac{19}{14}

\frac{x - 3}{x + 1} \geq 0

3 n + 4 n = - 14

p^{4} - q^{4}

8 x^{2} - 22 x + 5

x (x + 1)^{2} (x - 7) < 0

What are the solutions to the equation 7x^2-3x+1=0 ?
The solutions to the equation 7x^2-3x+1=0 are x= 3/14+i(sqrt(19))/(14),x= 3/14-i(sqrt(19))/(14)
Find the zeros of 7x^2-3x+1=0
The zeros of 7x^2-3x+1=0 are x= 3/14+i(sqrt(19))/(14),x= 3/14-i(sqrt(19))/(14)