What are the solutions to the equation 7x^2-5x+1=0 ?

The solutions to the equation 7x^2-5x+1=0 are x= 5/14+i(sqrt(3))/(14),x= 5/14-i(sqrt(3))/(14)

Find the zeros of 7x^2-5x+1=0

The zeros of 7x^2-5x+1=0 are x= 5/14+i(sqrt(3))/(14),x= 5/14-i(sqrt(3))/(14)

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7 x^{2} - 5 x + 1 = 0

x = \frac{5}{14} + i \frac{3}{14}, x = \frac{5}{14} - i \frac{3}{14}

Solution steps

7 x^{2} - 5 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

Simplify

(- 5)^{2} - 4 \cdot 7 \cdot 1 : 3 i

x_{1, 2} = \frac{- ( - 5 ) \pm 3 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 3 i}{2 \cdot 7}, x_{2} = \frac{- ( - 5 ) - 3 i}{2 \cdot 7}

x = \frac{- ( - 5 ) + 3 i}{2 \cdot 7} : \frac{5}{14} + i \frac{3}{14}

x = \frac{- ( - 5 ) - 3 i}{2 \cdot 7} : \frac{5}{14} - i \frac{3}{14}

The solutions to the quadratic equation are:

x = \frac{5}{14} + i \frac{3}{14}, x = \frac{5}{14} - i \frac{3}{14}

\frac{20 ab}{45 a ^{2} b ^{3}}

1 - (- 2)

\frac{4}{∣ x ∣ - 1} \leq ∣ x ∣ - 1

(3 x - 4)^{2} = 18

a^{3} - 1

What are the solutions to the equation 7x^2-5x+1=0 ?
The solutions to the equation 7x^2-5x+1=0 are x= 5/14+i(sqrt(3))/(14),x= 5/14-i(sqrt(3))/(14)
Find the zeros of 7x^2-5x+1=0
The zeros of 7x^2-5x+1=0 are x= 5/14+i(sqrt(3))/(14),x= 5/14-i(sqrt(3))/(14)