What are the solutions to the equation 9x^2-4x-1=0 ?

The solutions to the equation 9x^2-4x-1=0 are x=(2+sqrt(13))/9 ,x=(2-sqrt(13))/9

Find the zeros of 9x^2-4x-1=0

The zeros of 9x^2-4x-1=0 are x=(2+sqrt(13))/9 ,x=(2-sqrt(13))/9

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9 x^{2} - 4 x - 1 = 0

x = \frac{2 + 13}{9}, x = \frac{2 - 13}{9}

Decimal Notation

x = 0.62283 \dots, x = - 0.17839 \dots

Solution steps

9 x^{2} - 4 x - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 9 (- 1) = 213

x_{1, 2} = \frac{- ( - 4 ) \pm 2 13}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 13}{2 \cdot 9}, x_{2} = \frac{- ( - 4 ) - 2 13}{2 \cdot 9}

x = \frac{- ( - 4 ) + 2 13}{2 \cdot 9} : \frac{2 + 13}{9}

x = \frac{- ( - 4 ) - 2 13}{2 \cdot 9} : \frac{2 - 13}{9}

The solutions to the quadratic equation are:

x = \frac{2 + 13}{9}, x = \frac{2 - 13}{9}

(2 x^{2})^{2} (\frac{1}{2} y^{2})^{2}

- (9 x)^{0}

e^{- \frac{1}{x ^{7}}}

X^{3} + 27

4 p = 4

What are the solutions to the equation 9x^2-4x-1=0 ?
The solutions to the equation 9x^2-4x-1=0 are x=(2+sqrt(13))/9 ,x=(2-sqrt(13))/9
Find the zeros of 9x^2-4x-1=0
The zeros of 9x^2-4x-1=0 are x=(2+sqrt(13))/9 ,x=(2-sqrt(13))/9