What are the solutions to the equation 7x^2-6x+1=0 ?

The solutions to the equation 7x^2-6x+1=0 are x=(3+sqrt(2))/7 ,x=(3-sqrt(2))/7

Find the zeros of 7x^2-6x+1=0

The zeros of 7x^2-6x+1=0 are x=(3+sqrt(2))/7 ,x=(3-sqrt(2))/7

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7 x^{2} - 6 x + 1 = 0

x = \frac{3 + 2}{7}, x = \frac{3 - 2}{7}

Decimal Notation

x = 0.63060 \dots, x = 0.22654 \dots

Solution steps

7 x^{2} - 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

(- 6)^{2} - 4 \cdot 7 \cdot 1 = 22

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

Separate the solutions

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

x = \frac{- ( - 6 ) + 2 2}{2 \cdot 7} : \frac{3 + 2}{7}

x = \frac{- ( - 6 ) - 2 2}{2 \cdot 7} : \frac{3 - 2}{7}

The solutions to the quadratic equation are:

x = \frac{3 + 2}{7}, x = \frac{3 - 2}{7}

x^{3} + 2 x^{2} + 3 x + 6

∣ 2 x - 9 ∣ = 19

\frac{2 x h + h ^{2}}{h}

\frac{5 - x}{12} = - 7

49 x^{2} + 14 x + 1 > 0

What are the solutions to the equation 7x^2-6x+1=0 ?
The solutions to the equation 7x^2-6x+1=0 are x=(3+sqrt(2))/7 ,x=(3-sqrt(2))/7
Find the zeros of 7x^2-6x+1=0
The zeros of 7x^2-6x+1=0 are x=(3+sqrt(2))/7 ,x=(3-sqrt(2))/7