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

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

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

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

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

x = - \frac{- 1 + 7}{6}, x = \frac{1 + 7}{6}

Decimal Notation

x = - 0.27429 \dots, x = 0.60762 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

x = \frac{- 2 + 2 7}{2 ( - 6 )} : - \frac{- 1 + 7}{6}

x = \frac{- 2 - 2 7}{2 ( - 6 )} : \frac{1 + 7}{6}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 7}{6}, x = \frac{1 + 7}{6}

19 6^{\frac{1}{2}}

\frac{4 x ^{2}}{5 y} \cdot \frac{7 y}{12 x ^{4}}

42 m c + 36 m d - 7 n^{2} c - 6 n^{2} d

(a)^{2}

(3 x)^{2}

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