What are the solutions to the equation -9x^2-3x+1=0 ?

The solutions to the equation -9x^2-3x+1=0 are x=-(1+sqrt(5))/6 ,x=(sqrt(5)-1)/6

Find the zeros of -9x^2-3x+1=0

The zeros of -9x^2-3x+1=0 are x=-(1+sqrt(5))/6 ,x=(sqrt(5)-1)/6

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- 9 x^{2} - 3 x + 1 = 0

x = - \frac{1 + 5}{6}, x = \frac{5 - 1}{6}

Solution steps

- 9 x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 9 ) \cdot 1}{2 ( - 9 )}

(- 3)^{2} - 4 (- 9) \cdot 1 = 35

x_{1, 2} = \frac{- ( - 3 ) \pm 3 5}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 5}{2 ( - 9 )}, x_{2} = \frac{- ( - 3 ) - 3 5}{2 ( - 9 )}

x = \frac{- ( - 3 ) + 3 5}{2 ( - 9 )} : - \frac{1 + 5}{6}

x = \frac{- ( - 3 ) - 3 5}{2 ( - 9 )} : \frac{5 - 1}{6}

The solutions to the quadratic equation are:

x = - \frac{1 + 5}{6}, x = \frac{5 - 1}{6}

9 q^{2} - 11 q + 2 = - 3 q^{2}

- x^{2} + 1942 x - 0.0588 = 0

6 - \frac{p ^{2}}{8} = - 4

\frac{d}{- 12} = 144

6125

What are the solutions to the equation -9x^2-3x+1=0 ?
The solutions to the equation -9x^2-3x+1=0 are x=-(1+sqrt(5))/6 ,x=(sqrt(5)-1)/6
Find the zeros of -9x^2-3x+1=0
The zeros of -9x^2-3x+1=0 are x=-(1+sqrt(5))/6 ,x=(sqrt(5)-1)/6