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

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

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

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

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

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

Decimal Notation

x = 2.82287 \dots, x = 0.17712 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 2 \cdot 1 = 27

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{6 x y ^{6}}{4 x y ^{5}}

2 x^{4} y - 3 x^{2} y

x^{4} + 10 x^{2} + 21

8 x^{2} - 14 x + 3

\frac{5}{3} x - \frac{1}{6} = x + \frac{1}{4}

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