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

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

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

The zeros of quadraticformula 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}

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}

x^{2} - (3 m - 2) x + (m^{2} - 1) = 0

(x - 4)^{2} = 14 - x (2 x + 1)

5 n^{2} = 15 - 8 n

x^{2} + 8 x^{2} + 16 = 0

v^{2} + v - 6 = 0

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